Solving Quadratic Equations by Factoring
Once a quadratic equals zero and is factored, the zero-product rule hands you the solutions directly.
Reading the concept isn't enough. The score comes from practicing the real Bluebook-style format and finding the exact mistakes you keep making.
What the SAT tests here
First get everything on one side so the equation equals zero. Factor, then set each factor (not each term) equal to zero and solve.
How to solve one, step by step
Example: solve $x^2 - x - 6 = 0$.
- Factor: $(x - 3)(x + 2) = 0$.
- Set each factor to zero: $x = 3$ or $x = -2$.
The mistakes that cost points
- Not setting equal to zero first. The zero-product rule only works when one side is $0$.
- Setting each term to zero. Set each factor to zero, not each term.
Practice questions
Try these the way you would on test day, then open the solution to check your method.
Easy
Which of the following is equivalent to $2x^2 + 7x + 3$?
- A$(2x + 3)(x + 1)$
- B$(x + 3)(x + 1)$
- C$(2x + 1)(x + 3)$
- D$(2x + 7)(x + 3)$
Show solution
Answer: C, $(2x + 1)(x + 3)$. The leading coefficient is $2$, so factor by grouping. Multiply $a \times c = 2 \times 3 = 6$. Find two numbers that multiply to $6$ and add to $7$: $6$ and $1$. Rewrite: $2x^2 + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)$.
Medium
Which of the following is equivalent to $3x^2 - 10x - 8$?
- A$(x - 4)(x + 2)$
- B$(3x + 2)(x - 4)$
- C$(3x - 2)(x + 4)$
- D$(3x - 4)(x + 2)$
Show solution
Answer: B, $(3x + 2)(x - 4)$. Multiply $a \times c = 3 \times (-8) = -24$. Find two numbers that multiply to $-24$ and add to $-10$: $-12$ and $2$. Rewrite: $3x^2 - 12x + 2x - 8 = 3x(x - 4) + 2(x - 4) = (3x + 2)(x - 4)$.
Hard
The expression $6x^2 + kx - 10$ can be factored as $(2x + 5)(3x - 2)$. What is the value of $k$?
- A$-11$
- B$11$
- C$-7$
- D$7$
Show solution
Answer: B, $11$. Expand $(2x + 5)(3x - 2)$: $= 6x^2 - 4x + 15x - 10 = 6x^2 + 11x - 10$. So $k = 11$.
Find your exact gaps
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