Solving Quadratic Equations by Factoring
The idea behind it
If two numbers multiply to $0$, then at least one of them must be $0$. Factoring works because of that one fact. Get the equation equal to $0$, factor, and set each factor equal to $0$.
So the equation has to equal $0$ before you factor. From $x(x - 4) = 21$ you can't say that $x = 21$. First rewrite it as $x^2 - 4x - 21 = 0$.
Factoring is usually the fastest way to solve a quadratic on the SAT. When students lose points, it is because they factored before the equation equaled $0$, divided both sides by $x$ and lost a solution, or made a sign slip reading the solutions off the factors.
Three examples
Moving everything to one side
Solve $x(x - 4) = 21$.
- Expand and move $21$ over: $x^2 - 4x - 21 = 0$.
- Factor: two numbers that multiply to $-21$ and add to $-4$ are $-7$ and $3$. So $(x - 7)(x + 3) = 0$.
- $x = 7$ or $x = -3$.
A leading coefficient that is not $1$
Solve $2x^2 - 5x - 3 = 0$.
- Multiply $a \cdot c = 2 \cdot (-3) = -6$. Two numbers that multiply to $-6$ and add to $-5$ are $-6$ and $1$.
- Split the middle term: $2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3)$.
- $x = -\dfrac{1}{2}$ or $x = 3$.
A word problem
A rectangle is $3$ units longer than it is wide, and its area is $40$. Find the width.
- $w(w + 3) = 40$, so $w^2 + 3w - 40 = 0$.
- Factor: $(w + 8)(w - 5) = 0$, so $w = -8$ or $w = 5$.
- A width cannot be negative, so the width is $5$.
What you will see
| You see | Do this |
|---|---|
| $(x - 3)(x + 5) = 0$ | $x - 3 = 0$ or $x + 5 = 0$, so $x = 3$ or $x = -5$. The signs flip. |
| $x^2 + 2x = 15$ | Move $15$ over first: $x^2 + 2x - 15 = 0$, then factor. |
| $x^2 = 9x$ | Don't divide by $x$. Write $x^2 - 9x = 0$, so $x(x - 9) = 0$: $x = 0$ or $9$. |
| Leading coefficient not $1$ | Factor by grouping or check the choices: $2x^2 - 5x - 3 = (2x + 1)(x - 3)$. |
| A perfect square, like $4x^2 + 12x + 9$ | $(2x + 3)^2 = 0$ has exactly one solution. |
| Area or a product in a word problem | Write the equation, set it equal to $0$, factor, and keep only the positive answer. |
Mistakes I see often
- Factoring before the equation equals $0$. The zero product rule only works with $0$ on one side.
- Dividing by $x$. In $x^2 = 9x$, dividing by $x$ loses the solution $x = 0$.
- Keeping the signs from the factors. $(x + 5) = 0$ gives $x = -5$, not $5$.
- Missing a common factor. In $3x^2 - 12 = 0$, take out the $3$ first: $3(x^2 - 4) = 0$.
- Keeping a negative length. In area and distance problems, throw away the negative solution.
Practice questions
Get each equation equal to $0$ before you factor. Then check your answers in the solutions.
Easy
What are the solutions to $(x - 3)(x + 5) = 0$?
- A$x = -3$ and $x = 5$
- B$x = 3$ and $x = -5$
- C$x = 3$ and $x = 5$
- D$x = -15$ only
Show solution
Answer: B. Set each factor equal to $0$: $x - 3 = 0$ gives $x = 3$, and $x + 5 = 0$ gives $x = -5$. Choice A keeps the signs from the factors.
EasyStudent-produced response
What is the greater solution to $x^2 - 7x + 12 = 0$?
Enter your answer.
Show solution
Answer: $4$. $(x - 3)(x - 4) = 0$, so $x = 3$ or $x = 4$. The greater solution is $4$.
Medium
What are the solutions to $x^2 + 2x = 15$?
- A$x = 3$ and $x = -5$
- B$x = -3$ and $x = 5$
- C$x = 15$ and $x = -2$
- D$x = 0$ and $x = -2$
Show solution
Answer: A. $x^2 + 2x - 15 = 0$ factors as $(x + 5)(x - 3) = 0$, so $x = 3$ or $x = -5$. Choice D sets $x^2 + 2x = 0$, ignoring the $15$.
Medium
What are the solutions to $2x^2 - 5x - 3 = 0$?
- A$x = \dfrac{1}{2}$ and $x = -3$
- B$x = -1$ and $x = 3$
- C$x = -\dfrac{1}{2}$ and $x = 3$
- D$x = -2$ and $x = 3$
Show solution
Answer: C. $2x^2 - 5x - 3 = (2x + 1)(x - 3)$. Setting each factor to $0$: $x = -\dfrac{1}{2}$ or $x = 3$. Choice A has both signs reversed.
Medium
What are all the solutions to $x^2 = 9x$?
- A$x = 9$ only
- B$x = -9$ and $x = 9$
- C$x = 0$ and $x = -9$
- D$x = 0$ and $x = 9$
Show solution
Answer: D. $x^2 - 9x = 0$, so $x(x - 9) = 0$ and $x = 0$ or $x = 9$. Choice A divides both sides by $x$, which loses the solution $0$.
MediumStudent-produced response
What is the positive solution to $3x^2 - 12 = 0$?
Enter your answer.
Show solution
Answer: $2$. $3(x^2 - 4) = 0$, so $3(x - 2)(x + 2) = 0$. The solutions are $2$ and $-2$, and the positive one is $2$.
Hard
What is the positive solution to $x(x - 4) = 21$?
- A$7$
- B$21$
- C$25$
- D$3$
Show solution
Answer: A. $x^2 - 4x - 21 = 0$ factors as $(x - 7)(x + 3) = 0$, so $x = 7$ or $x = -3$. Choices B and C set each factor equal to $21$, which only works when the product is $0$.
HardStudent-produced response
A rectangle is $3$ units longer than it is wide. Its area is $40$ square units. What is the width of the rectangle, in units?
Enter your answer.
Show solution
Answer: $5$. $w(w + 3) = 40$ gives $w^2 + 3w - 40 = 0$, so $(w + 8)(w - 5) = 0$. A width must be positive, so $w = 5$.
HardStudent-produced response
One solution of $x^2 + kx + 18 = 0$ is $x = -3$, where $k$ is a constant. What is the value of $k$?
Enter your answer.
Show solution
Answer: $9$. Substitute $x = -3$: $9 - 3k + 18 = 0$, so $3k = 27$ and $k = 9$. Check: $x^2 + 9x + 18 = (x + 3)(x + 6)$.
Hard
Which statement is true about the equation $4x^2 + 12x + 9 = 0$?
- AIt has two solutions, $-\dfrac{3}{2}$ and $\dfrac{3}{2}$.
- BIt has exactly one solution, $-\dfrac{3}{2}$.
- CIt has exactly one solution, $\dfrac{3}{2}$.
- DIt has no real solutions.
Show solution
Answer: B. $4x^2 + 12x + 9 = (2x + 3)^2$, so $2x + 3 = 0$ and $x = -\dfrac{3}{2}$. A perfect square has only one solution.
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