Solving polynomial equations that are already written as a product equal to zero. Includes factored form and the zero-product property, setting each factor equal to zero, repeated factors giving a single solution, equations with three or more factors, sketching a parabola from its factored form, and using a factored quadratic model of an arch.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 10 — Polynomials and Factoring
Solving Quadratic Equations in Factored Form
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.4 Solving Quadratic Equations in Factored Form §10.4, pp. 588-594 — the lesson these objectives are drawn from
Warm-up
Lesson 10.2 multiplied binomials together. This lesson starts from the product and works backwards, but only when the product is nought.
Discussion prompt
Two numbers multiply to give nought. What must be true of them? Now answer the same question if they multiply to give twelve.
Hint: Try to find two non-zero numbers whose product is nought.
Answer:
At least one of them must be nought. There is no pair of non-zero numbers whose product is nought, because multiplying two non-zero quantities always gives something non-zero.
For twelve there is no such conclusion at all: three and four work, so do two and six, and so do countless fractions. That difference is exactly why equations have to be set equal to nought before they are factored, and it is the whole idea of this lesson.
Concept
A polynomial is in factored form if it is written as a product of two or more factors. If such a product equals nought, then at least one of the factors must equal nought.
zero-product property — If a and b are real numbers and their product is nought, then a is nought or b is nought. If the product of two factors is nought, at least one of the factors must be nought.
That is what makes a factored equation easy to solve.
Figure (svg): The zero-product property stated and illustrated
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.4 Solving Quadratic Equations in Factored Form §10.4, pp. 588-588
Section
Section 1
Concept
To solve a factored equation, set each factor equal to nought and solve the resulting equations separately. Every solution of the original is a solution of one of them, and there are no others.
The zero-product property is what guarantees no solutions are missed.
Figure (svg): A factored equation split into two simple equations
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.4 Solving Quadratic Equations in Factored Form §10.4, pp. 588-588 — the Zero-Product Property and Example 1
Picture it
One equation each.
Figure (svg): A factored equation split into two simple equations
The word between the branches is or rather than and: a solution needs only one factor to vanish, not both. Reading it as and would produce no solutions at all here.
Worked example
This is Example 1 from the textbook.
\[ \text{Solve } (x + 2)(x - 3) = 0. \]
Apply the property
Why: One of the two factors must be nought.
\[ x + 2 = 0 \text{ or } x - 3 = 0 \]
Solve the first
Why: Subtract two.
\[ x = -2 \]
Solve the second
Why: Add three.
\[ x = 3 \]
Collect
Why: Both are solutions.
\[ x = -2, \; 3 \]
Figure (svg): A factored equation split into two simple equations
\[ x = -2 \text{ and } x = 3 \]
Verify: substitute both back
Why: At negative two the first factor is nought, so the product is nought whatever the second factor is; at three the second factor is nought. Each solution works by killing one factor, which is exactly what the property describes.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.4 Solving Quadratic Equations in Factored Form §10.4, pp. 588-588
Matching
The solution is the opposite of the constant.
Match the pairs
Why: A plus in the factor gives a negative solution and a minus gives a positive one. The bare x is its own case: it is already solved, and nought is a legitimate solution rather than an absence of one.
Worked example
Guided Practice 1 to 3.
\[ \text{Solve } (x + 1)(x + 3) = 0, \; x(x - 2) = 0 \text{ and } (x - 5)(x + 7) = 0. \]
Take the first
Why: Both factors are sums.
\[ x = -1, \; - 3 \]
Take the second
Why: The first factor is just x.
\[ x = 0 \text{ or } x - 2 = 0 \]
Finish the second
Why: Nought is a perfectly good solution.
\[ x = 0, \; 2 \]
Take the third
Why: One sum and one difference.
\[ x = 5, \; - 7 \]
Figure (svg): A factored equation split into two simple equations
\[ -1, -3; \quad 0, 2; \quad 5, -7 \]
Verify: check the sign flip in each solution
Why: The factor x plus one gives x equal to negative one, and x minus five gives x equal to five — each solution is the opposite of the constant in its factor. That reversal is worth noticing, because reading the solutions straight off the factors without flipping is a common slip.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.4 Solving Quadratic Equations in Factored Form §10.4, pp. 588-588
Trap
\[ (x + 2)(x - 3) = 6 \;\Longrightarrow\; x + 2 = 6 \text{ or } x - 3 = 6 \]
Set each factor equal to the number on the right
Why: The method worked for nought, so it was applied to six.
It gives x equal to four or nine, and neither works: at four the product is six times one, which is six — that one happens to work by luck — but at nine it is eleven times six, which is sixty-six. There are countless ways two numbers can multiply to six, so no single factor is forced to any value.
\[ (x + 2)(x - 3) = 6 \;\to\; x^2 - x - 12 = 0 \]
Expand, move everything to one side, then factor again
Why: The property needs a zero on the right.
Nought is the only number with this property, which is why standard form always puts it there.
Faded example
One equation per factor.
Fill in the blanks
(x + 2)(x - 3) = 0 \;\Longrightarrow\; x + 2 = 0 \text0 x - 3 = ___
Why: Both factors are set to nought, not to anything else, because nought is the only value the property says anything about. Setting them to the number on the right of the original equation is the classic misuse.
Elimination
Look at the right-hand side.
Eliminate the wrong options
To which equation can the zero-product property be applied directly?
Survives elimination: A
Why: The property needs a product on one side and nought on the other. Option B is the case worth dwelling on, since it looks almost identical and the method silently produces wrong answers.
Socratic
Six seems like it should work too.
Discussion prompt
Explain why a product equal to nought forces a factor to be nought, while a product equal to six forces nothing. Then say what that means for how equations must be arranged before factoring.
Hint: Count the ways each product can be achieved.
Answer:
Multiplying two non-zero numbers always gives a non-zero result, so the only way to reach nought is for one of them to be nought already — there is exactly one route. For six there are endlessly many routes: one and six, two and three, twelve and a half, and so on, so knowing the product is six tells you nothing about either factor on its own.
That is why every equation must be rearranged so that one side is nought before factoring is any use. It is the same reason standard form puts a nought on the right in Lessons 9.5 and 9.6, and it explains why the very first step of every factoring method in the rest of this chapter is to move all the terms to one side.
Section
Section 2
Concept
A repeated factor is one that appears twice or more. Setting it equal to nought gives the same equation each time, so the equation has a single solution rather than two.
\[ (x + 5)^2 = 0 \;\Longrightarrow\; x + 5 = 0 \]
This is the one-solution case of Lesson 9.7.
Figure (svg): A repeated factor giving a single solution
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.4 Solving Quadratic Equations in Factored Form §10.4, pp. 589-589 — Example 2, Solve a Repeated-Factor Equation
Picture it
Both give the same equation.
Figure (svg): A repeated factor giving a single solution
Nothing is lost by writing only one solution: the second branch was the same equation and would have produced the same number. Listing it twice would be a repetition rather than an extra answer.
Worked example
This is Example 2 from the textbook.
\[ \text{Solve } (x + 5)^2 = 0. \]
Recognise the repetition
Why: The square is the factor twice.
\[ (x + 5) (x + 5) = 0 \]
Set the factor to nought
Why: Both branches are identical.
\[ x + 5 = 0 \]
Solve
Why: Subtract five.
\[ x = -5 \]
Report one solution
Why: The branches coincide.
\[ x = -5 \]
Figure (svg): A repeated factor giving a single solution
\[ x = -5 \]
Verify: substitute it back
Why: Negative five plus five is nought, and nought squared is nought, so the equation holds. The check also shows why there is only one answer: no other value makes that bracket vanish.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.4 Solving Quadratic Equations in Factored Form §10.4, pp. 589-589
Sorting
Count the distinct factors.
Sort into buckets
Sort each equation by its number of distinct solutions.
Counting distinct factors rather than counting brackets is what gives the right answer. A squared bracket is two factors but only one distinct one.
Worked example
Guided Practice 4 to 6, one with a coefficient.
\[ \text{Solve } (x - 4)^2 = 0, \; (x + 6)^2 = 0 \text{ and } (2x - 5)^2 = 0. \]
Take the first
Why: Set the factor to nought.
\[ x = 4 \]
Take the second
Why: The constant is added.
\[ x = -6 \]
Take the third
Why: A coefficient to divide out.
\[ 2 x = 5 \]
Finish the third
Why: Divide by two.
\[ x = \tfrac{5}{2} \]
Figure (svg): A repeated factor giving a single solution
\[ x = 4, \quad x = -6, \quad x = \tfrac{5}{2} \]
Verify: check the fractional one
Why: Two times five halves is five, and five minus five is nought, so the squared factor is nought. A fractional solution is as legitimate as a whole-number one, and it arises whenever the factor has a coefficient other than one.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.4 Solving Quadratic Equations in Factored Form §10.4, pp. 589-589
Trap
\[ (x + 5)^2 = 0 \;\Longrightarrow\; x + 5 = \pm 0 \;\Longrightarrow\; x = -5 \text{ or } x = -5 \]
Write plus-or-minus, since a square root normally gives two values
Why: Lesson 9.2 always wrote both signs.
The plus-or-minus is harmless here but pointless: positive nought and negative nought are the same number, so both branches give the same solution. Reporting two solutions would be wrong.
\[ (x + 5)^2 = 0 \;\Longrightarrow\; x + 5 = 0 \;\Longrightarrow\; x = -5 \]
Note that nought is its own opposite, so only one branch exists
Why: The two solutions have coincided.
This is exactly Lesson 9.2's middle case, where d equals nought gives one solution.
Prediction
A quadratic with a repeated factor.
Predict first
How does the graph of y equal to (x plus five) squared meet the x-axis?
Correct: It touches the axis at one point and turns back.
\[ (x + 5)^2 = 0 \text{ only at } x = -5 \]
Why: The single solution is the only place the function is nought, and since the parabola opens upwards it must approach the axis, touch, and rise again. That touching point is the vertex, which is exactly the one-solution case from Lesson 9.7 where the discriminant is nought. A repeated factor and a discriminant of nought are two descriptions of the same situation.
Faded example
One extra step at the end.
Fill in the blanks
(2x - 5)^2 = 0 \;\to\; 2x - 5 = 0 \;\to\; 2x = 5 \;\to\; x = 5/2
Why: The coefficient has to be divided out after the factor is set to nought, not before. Doing it in that order keeps every step a simple linear equation.
Socratic
Quadratics usually have two.
Discussion prompt
Explain why a repeated factor gives a genuine single solution rather than a lost one. Then say how this connects to the discriminant of Lesson 9.7.
Hint: Ask what the second branch would have given.
Answer:
Both branches of the zero-product property are used, and both produce the same equation and the same number. Nothing has been lost — the two solutions have simply arrived at the same place, in the way that the two roots of x squared equals d meet when d reaches nought.
Expanding the squared factor gives a quadratic whose discriminant is exactly nought, which Lesson 9.7 said means one solution. So the repeated factor and the zero discriminant are the same fact seen from two sides, and later work often calls this a double root — one value that counts twice for some purposes and once as an answer.
Section
Section 3
Concept
If a product of several factors is nought, at least one of them is nought. Setting each factor equal to nought gives one simple equation per factor.
\[ (2x + 1)(3x - 2)(x + 1) = 0 \]
A cubic in factored form is no harder than a quadratic.
Figure (svg): A product of three factors giving three solutions
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.4 Solving Quadratic Equations in Factored Form §10.4, pp. 589-589 — Example 3, Solve a Factored Cubic Equation
Picture it
One equation each.
Figure (svg): A product of three factors giving three solutions
The degree of the expanded polynomial would be three, and it has three solutions — one more than any quadratic can have. Factored form makes that visible without any expansion.
Worked example
This is Example 3 from the textbook.
\[ \text{Solve } (2x + 1)(3x - 2)(x + 1) = 0. \]
Set each factor to nought
Why: Three equations.
\[ 2 x + 1 = 0, \; 3 x - 2 = 0, \; x + 1 = 0 \]
Solve the first
Why: Subtract, then divide.
\[ x = -\tfrac{1}{2} \]
Solve the second
Why: Add, then divide.
\[ x = \tfrac{2}{3} \]
Solve the third
Why: Subtract one.
\[ x = -1 \]
Figure (svg): A product of three factors giving three solutions
\[ x = -\tfrac{1}{2}, \; \tfrac{2}{3}, \; -1 \]
Verify: check one solution
Why: At x equal to two thirds the second factor is two minus two, which is nought, so the whole product is nought however large the other two factors are. Each solution works by killing exactly one factor, which is worth confirming once.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.4 Solving Quadratic Equations in Factored Form §10.4, pp. 589-589
Faded example
Isolate, then divide.
Fill in the blanks
3x - 2 = 0 \;\to\; 3x = 2 \;\to\; x = 2/3
Why: Two steps rather than one, because the factor has a coefficient. Reading the solution straight off as two would skip the division and give a value that fails the check.
Worked example
Guided Practice 9 and 10, where one factor is squared.
\[ \text{Solve } (2x - 1)(x + 8)^2 = 0 \text{ and } (y - 3)^2(3y + 2) = 0. \]
Take the first equation
Why: Two distinct factors, one squared.
\[ 2x-1 = 0 \text{ or } x+8 = 0 \]
Solve them
Why: A fraction and a whole number.
\[ x = \tfrac{1}{2}, \; -8 \]
Take the second equation
Why: Again two distinct factors.
\[ y-3 = 0 \text{ or } 3y+2 = 0 \]
Solve them
Why: A whole number and a fraction.
\[ y = 3, \; -\tfrac{2}{3} \]
Figure (svg): A product of three factors giving three solutions
\[ x = \tfrac{1}{2}, -8; \qquad y = 3, -\tfrac{2}{3} \]
Verify: count distinct factors rather than brackets
Why: Each equation is cubic when expanded, but each has only two distinct factors and therefore two solutions. The repeated factor contributes one solution, not two, which is why the count is two rather than three.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.4 Solving Quadratic Equations in Factored Form §10.4, pp. 589-589
Error analysis
The student solved a factored cubic.
Annotate
On: \( \begin{aligned} (2x + 1)(3x - 2)(x + 1) &= 0 \\ 2x + 1 &= 0 \;\Longrightarrow\; x = -1 \\ 3x - 2 &= 0 \;\Longrightarrow\; x = -2 \\ x + 1 &= 0 \;\Longrightarrow\; x = 1 \end{aligned} \)
Both errors come from reading a solution off a factor rather than solving the factor's equation. Writing each factor as its own line of working, however trivial it looks, is what keeps the coefficient and the sign from being skipped.
Translation
Each needs its own small equation.
Match the pairs
Why: Three of the four need a division as well as a sign change, and the fractions that result are perfectly ordinary answers. A factor with a coefficient rarely gives a whole-number solution.
Elimination
For the equation (2x - 1)(x + 8) squared = 0.
Eliminate the wrong options
How many distinct solutions does it have?
Survives elimination: A
Why: Counting distinct factors gives two: two x minus one and x plus eight. The squared factor contributes a single value even though it appears twice, which is the same situation as the repeated factor of the previous section.
Socratic
A quadratic can have at most two.
Discussion prompt
Explain why a product of three linear factors can be nought at three different values. Then say what that implies about the shape of its graph.
Hint: Each factor vanishes somewhere different.
Answer:
Each linear factor is nought at exactly one value of x, and if the three factors are different those three values are different. Any one of them makes the whole product nought, so all three are solutions — and there can be no more, because away from those three values every factor is non-zero and so is their product.
The graph must therefore cross the x-axis three times, which a parabola cannot do since it turns only once. A curve crossing three times has to turn at least twice, so a cubic graph has two turning points rather than one — which is why Lesson 10.8's cubics look quite different from anything in Chapter 9.
Section
Section 4
Concept
The x-intercepts of a function in factored form are the solutions of the corresponding equation. The vertex lies halfway between them, and substituting that value gives its height.
Three points and a direction are enough for a sketch.
Figure (svg): A parabola sketched from its factored form
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.4 Solving Quadratic Equations in Factored Form §10.4, pp. 590-590 — Example 4, Graph a Factored Equation
Picture it
All three from the factors.
Figure (svg): A parabola sketched from its factored form
No table is needed and no formula for the vertex either. Factored form gives up more about the graph than standard form does, which is one reason factoring is worth the effort.
Worked example
This is Example 4 from the textbook.
\[ \text{Sketch the graph of } y = (x - 3)(x + 2). \]
Find the intercepts
Why: Solve the factored equation.
\[ 3 \text{ and } -2 \]
Average them
Why: The axis of symmetry.
\[ x = \tfrac{1}{2} \]
Substitute back
Why: Both brackets at a half.
\[ \left(-\tfrac{5}{2}\right)\left(\tfrac{5}{2}\right) \]
Simplify
Why: The vertex's height.
\[ -\tfrac{25}{4} \]
Figure (svg): A parabola sketched from its factored form
\[ \text{intercepts } 3, -2; \quad \text{vertex } \left(\tfrac{1}{2}, -\tfrac{25}{4}\right) \]
Verify: check the axis against the standard-form formula
Why: Expanding gives x squared minus x minus six, so negative b over two a is one over two, agreeing with the average of the intercepts. Two independent routes to the axis is a good check, and averaging is the quicker of the two.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.4 Solving Quadratic Equations in Factored Form §10.4, pp. 590-590
Faded example
The vertex sits between them.
Fill in the blanks
\text0.5 3 \text-6.25 -2 \;\to\; x = \dfrac______ = ___ \;\to\; y = ___
Why: The average locates the axis by symmetry, with no formula required. Substituting it back into the factored form is easier than expanding first, since both brackets are simple.
Worked example
Guided Practice 11 to 13.
\[ \text{Find the intercepts and vertex of } y = x(x - 2), \; y = (x + 4)(x - 5) \text{ and } y = (x - 1)(x + 6). \]
Take the first
Why: Intercepts at nought and two.
\[ \text{vertex } (1, -1) \]
Take the second
Why: Intercepts at negative four and five.
\[ \text{axis } x = \tfrac{1}{2} \]
Finish the second
Why: Substitute a half.
\[ \left(\tfrac{1}{2}, -\tfrac{81}{4}\right) \]
Take the third
Why: Intercepts at one and negative six.
\[ \left(-\tfrac{5}{2}, -\tfrac{49}{4}\right) \]
Figure (svg): The vertex located as the midpoint of the two x-intercepts
\[ (1, -1), \quad \left(\tfrac{1}{2}, -\tfrac{81}{4}\right), \quad \left(-\tfrac{5}{2}, -\tfrac{49}{4}\right) \]
Verify: notice all three vertices are below the axis
Why: Each of these has two distinct intercepts, so the curve must dip below the axis between them — an upward parabola crossing twice has its lowest point in between. A positive vertex height would contradict having two real intercepts.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.4 Solving Quadratic Equations in Factored Form §10.4, pp. 590-590
Trap
\[ y = (x - 3)(x + 2) \;\Longrightarrow\; \text{intercepts at } -3 \text{ and } 2 \]
Copy the constants out of the brackets
Why: The numbers three and two are right there.
Each intercept is where its factor is nought, so x minus three vanishes at positive three and x plus two vanishes at negative two. The signs are the opposite of those written in the brackets.
\[ \text{intercepts at } 3 \text{ and } -2 \]
Solve each factor's equation rather than reading its constant
Why: One line each.
Substituting one intercept back into the function is a five-second confirmation.
Sorting
Which form gives each feature more easily.
Sort into buckets
Sort each feature by which form hands it over with less work.
Neither form wins outright, which is why converting between them is worth being able to do. Each hides exactly what the other displays.
Elimination
For y equal to (x - 3)(x + 2).
Eliminate the wrong options
What is the x-coordinate of the vertex?
Survives elimination: A
Why: The intercepts are three and negative two, whose average is a half. Option D is the slip that follows from reading the constants without flipping their signs, and it produces a plausible-looking number.
Socratic
It is stated as a fact.
Discussion prompt
Explain why the vertex's x-coordinate is the average of the two intercepts. Then say what happens to this method when there are no real intercepts.
Hint: Use the symmetry of the parabola.
Answer:
The parabola is symmetric about its axis, and the two intercepts are both at height nought, so they are mirror images of each other. Mirror images are equidistant from the axis, which puts the axis exactly at their midpoint — and the vertex sits on the axis by definition.
With no real intercepts there is nothing to average, so the method fails and the formula from Lesson 9.4 must be used instead. That is not a defect: a quadratic with no real roots cannot be written as a product of linear factors with real coefficients in the first place, so it would never appear in factored form here.
Section
Section 5
Concept
When a situation is modelled by a factored quadratic, the intercepts often have a direct physical meaning — the ends of an arch, the start and finish of a flight, the edges of a region.
A model written in factored form is written that way for a reason.
Figure (svg): A parabolic arch with its width and height marked
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.4 Solving Quadratic Equations in Factored Form §10.4, pp. 590-590 — Example 5, Use a Quadratic Model, on the arch
Picture it
Span and height from the factors.
Figure (svg): A parabolic arch with its width and height marked
Both quantities anyone would want to know — how wide and how tall — come out of the factored form in two short steps. Expanding it first would make both harder to see.
Worked example
This is Example 5 from the textbook.
\[ \text{An arch is modelled by } y = -0.15(x + 8)(x - 8). \text{ Find its base width and height.} \]
Find the intercepts
Why: Set each factor to nought.
\[ -8 \text{ and } 8 \]
Find the width
Why: The distance between them.
\[ 8 + 8 = 16 \]
Find the axis
Why: Average the intercepts.
\[ x = 0 \]
Find the height
Why: Substitute nought.
\[ -0.15(8) (-8) = 9.6 \]
Figure (svg): A parabolic arch with its width and height marked
\[ \text{width } 16 \text{ ft}, \quad \text{height } 9.6 \text{ ft} \]
Verify: check the sign of the leading coefficient
Why: The coefficient is negative, so the parabola opens downwards and the vertex is the highest point — which is what an arch requires. A positive coefficient would give a valley rather than an arch, and the height would be a minimum.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.4 Solving Quadratic Equations in Factored Form §10.4, pp. 590-590
Faded example
Two short steps from the factors.
Fill in the blanks
y = -0.15(x + 8)(x - 8): \quad \text16 = 8 - (-8) = 9.6, \quad \text___ = -0.15(8)(-8) = ___
Why: The width comes from the difference of the intercepts and the height from substituting their midpoint. Neither calculation needs the expanded form of the model.
Worked example
What the eight controls.
\[ \text{How would } y = -0.15(x + 10)(x - 10) \text{ differ from the original arch?} \]
Find the new intercepts
Why: Set each factor to nought.
\[ -10 \text{ and } 10 \]
Find the new width
Why: Twenty feet.
\[ 20 \]
Find the new height
Why: Substitute nought.
\[ -0.15(10) (-10) = 15 \]
Compare
Why: Wider and taller.
\[ 16 \to 20, \; 9.6 \to 15 \]
Figure (svg): A parabolic arch with its width and height marked
\[ \text{width } 20 \text{ ft}, \quad \text{height } 15 \text{ ft} \]
Verify: compare the two increases
Why: The width grew by a quarter and the height by more than half, because the height depends on the product of the two intercepts rather than on their difference. Widening an arch of this family makes it disproportionately taller, which is a real constraint on where such a shape can be used.
Trap
\[ \text{intercept } 8 \;\Longrightarrow\; \text{the arch is } 8 \text{ feet wide} \]
Report the positive intercept as the span
Why: It is the number that came out of the factor.
The arch stretches from negative eight to eight, so its width is the distance between them, which is sixteen. Eight is the distance from the centre to one foot.
\[ 8 - (-8) = 16 \text{ feet} \]
Take the distance between the two intercepts
Why: Both feet, not one.
This is the same doubling step that the bridge model needed in Lesson 9.5.
Hypothesis
The arch model has a negative 0.15 in front.
Predict first
What would happen if that number were changed to negative 0.3?
Correct: The arch would keep its width but double in height.
\[ -0.3(8)(-8) = 19.2 \]
Why: The intercepts come from the factors alone, so the feet stay at negative eight and eight and the span is unchanged. The height is the coefficient times the product of the bracket values, so doubling the coefficient doubles it to 19.2 feet. The coefficient controls how steep or shallow the arch is without moving where it meets the ground, which is exactly the freedom a designer needs.
Elimination
For a model with intercepts at -8 and 8.
Eliminate the wrong options
Which calculation gives the arch's height?
Survives elimination: A
Why: The height is the vertex's y-coordinate, and the vertex sits at the midpoint of the intercepts, which here is nought. Every wrong option is a horizontal measurement or a shape parameter rather than a vertical distance.
Socratic
Standard form is the usual way.
Discussion prompt
Say what a factored model makes obvious that an expanded one hides. Then say when you would want the expanded version instead.
Hint: Think about which numbers a designer cares about.
Answer:
The factors name the two places where the curve meets the ground, which for an arch are its feet — the single most important pair of numbers about it. Expanding the arch model gives negative 0.15 x squared plus 9.6, from which the span has to be recovered by solving, so the form actively hides what the designer most wants.
The expanded form is better when the y-intercept matters, since that is just the constant term, or when the model must be added to or subtracted from another one, since like terms only line up in standard form. Each form is worth having, and converting between them is what the rest of this chapter is about.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Factored form | Standard form | |
|---|---|---|
| x-intercepts | read off the factors | found with the quadratic formula |
| y-intercept | found by substituting zero | the constant term c |
| Axis of symmetry | average the intercepts | negative b over two a |
Each column hides what the other displays. That is the whole reason for learning to move between them, which the next four lessons are about.
Pattern
To solve any equation already written as a product equal to nought, these five moves cover it.
Step one is not a formality: applying the property to a product equal to anything other than nought produces answers that look reasonable and are not.
OpenStax Elementary Algebra 2e, §7.6 Quadratic Equations §7.6
Check
Flip the signs.
Check your understanding
Solve (x + 2)(x - 3) = 0.
Answer: A
Why: The factor x plus two is nought at negative two, and x minus three is nought at three. Each solution is the opposite of the constant in its factor.
Check
A repeated factor.
Check your understanding
How many distinct solutions has (x + 5) squared = 0?
Answer: A
Why: Both branches give the same equation, x plus five equals nought, so the two solutions coincide at negative five.
Check
The vertex is the midpoint.
Check your understanding
What is the vertex of y = (x - 3)(x + 2)?
Answer: A
Why: The intercepts are three and negative two, whose average is a half, and substituting a half gives negative twenty-five quarters.
Real world
This is the crater question from the lesson opener. The cross-section of a crater or an arch can be modelled by a quadratic written in factored form, whose intercepts are where the curve meets ground level.
Discussion prompt
An arch is modelled by y equal to negative 0.15 times (x + 8)(x - 8), with x and y in feet. Find its width at the base and its height, and then say what would change if the model were negative 0.15 times (x + 8)(x - 12).
Hint: The intercepts are the feet of the arch.
Answer:
\[ \text{intercepts } -8 \text{ and } 8 \;\Longrightarrow\; \text{width } 16 \text{ ft}, \quad \text{height } -0.15(8)(-8) = 9.6 \text{ ft} \]
With intercepts at negative eight and twelve the arch would be twenty feet wide, but it would no longer be symmetric about the vertical axis — its peak would sit at x equal to two, the average of the two feet.
Substituting two gives negative 0.15 times ten times negative ten, which is fifteen feet. So an arch of this family gets taller as it gets wider, and moving one foot without moving the other tilts the whole shape sideways. All three facts came from the factors without ever expanding the model, which is why arch and crater profiles are usually written this way.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
How should you solve (x + 2)(x - 3) = 6?
Correct: Expand, move the 6 across, and factor again.
\[ (x+2)(x-3) = 6 \;\to\; x^2 - x - 12 = 0 \;\to\; x = 4, -3 \]
Why: The zero-product property says nothing about a product equal to six, because there are countless pairs of numbers whose product is six — one and six, two and three, and endlessly many fractions — so no single factor is forced to any particular value. The first option happens to produce one correct answer by luck and one wrong one, which makes it more dangerous than an outright failure. Expanding gives x squared minus x minus six equals six, so x squared minus x minus twelve equals nought, which factors as x minus four times x plus three and gives the solutions four and negative three. Nought is the only number with this property, which is precisely why every method in this chapter begins by arranging for a zero on one side.
Explain it
They solved a factored equation and wrote the intercepts as the numbers inside the brackets.
Discussion prompt
In no more than four sentences, explain why the signs come out the opposite way. Then give them a check that takes five seconds.
Hint: What makes each bracket vanish?
Answer:
A usable answer: a solution is a value that makes one bracket equal nought, so for x minus three you need x to be three, and for x plus two you need x to be negative two. The sign in the bracket and the sign of the solution are always opposite, because you are undoing whatever the bracket does.
The check is to put your answer back into the bracket it came from and see whether it gives nought. If x plus two does not become nought when you substitute, that value is not a solution.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: The zero requirement is fixed by checking the right-hand side before doing anything else. Signs are fixed by writing each factor as its own small equation rather than reading a number off. Repeated factors are fixed by counting distinct factors rather than brackets. The vertex is fixed by averaging the intercepts and substituting back. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of a page write the zero-product property, and beside it write one product equal to nought and one equal to six, with a sentence under each saying what can and cannot be concluded. Underneath, solve a two-factor equation writing each factor's equation on its own line, and circle the sign change between the bracket's constant and the solution. In the middle, solve a repeated-factor equation and a three-factor equation side by side, and write the count of distinct solutions beside each with a one-line reason. Beneath that, take a factored quadratic and build its graph: mark both intercepts on an axis, average them on a number line to locate the vertex, substitute back for the height, and sketch the curve through all three points. In the lower corner, draw an arch, label its two feet and its peak, and work out the span and height from a factored model. Finally, in the margin, write which features factored form gives easily and which standard form gives easily.
Your averaged vertex should agree with negative b over two a if you expand the quadratic and check. If the two disagree, the intercepts were most likely read off with the wrong signs.
Recap
Five things, and the first is what makes all the others possible.
| If the question says | Your first move is |
|---|---|
| A product equals zero | Set each factor equal to zero |
| A product equals something else | Expand and rearrange first |
| A factor is squared | One solution, not two |
| Sketch a factored quadratic | Intercepts, then their midpoint |
| A model is given in factored form | Read the intercepts as physical points |
Lesson 10.5 begins the reverse of Lesson 10.2. Given a trinomial in standard form, the task is to find the two binomials that produce it — and once that is done, this lesson's property finishes the job of solving.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 10 Polynomials and Factoring — Lesson 10.4 Solving Quadratic Equations in Factored Form §10.4, pp. 588-594 — everything on these slides traces back here
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