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Difficulty: MediumSolving Quadratic Equations by Factoring

For each of the given quadratic equations, solve for xx by factoring. Match each quadratic equation on the left to its correct solution set on the right.

  • The equation 2x2+5x=32x^2 + 5x = 3The solution set {−3,12}\left\{-3, \frac{1}{2}\right\}
  • The equation 3x2−10x+8=03x^2 - 10x + 8 = 0The solution set {43,2}\left\{\frac{4}{3}, 2\right\}
  • The equation x(x+2)=15x(x + 2) = 15The solution set {−5,3}\left\{-5, 3\right\}

Answer

The equation 2x2+5x=32x^2 + 5x = 3 matches the solution set {−3,12}\left\{-3, \frac{1}{2}\right\}; the equation 3x2−10x+8=03x^2 - 10x + 8 = 0 matches the solution set {43,2}\left\{\frac{4}{3}, 2\right\}; and the equation x(x+2)=15x(x + 2) = 15 matches the solution set {−5,3}\left\{-5, 3\right\}.
Each equation matches its corresponding solution set through distributing terms if necessary, rewriting the equation in standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0, factoring the quadratic trinomial over the integers, and then using the zero product property to solve for xx.

Step-by-Step Solution

1
For the equation 2x2+5x=32x^2 + 5x = 3, rewrite in standard form ax2+bx+c=0ax^2 + bx + c = 0 by subtracting 3 from both sides to get 2x2+5x−3=02x^2 + 5x - 3 = 0.
The equation is rewritten as 2x2+5x−3=02x^2 + 5x - 3 = 0.
Before factoring a quadratic equation, all terms must be moved to one side so the other side is equal to zero.
2
Factor the trinomial 2x2+5x−32x^2 + 5x - 3 by grouping. Find two integers that multiply to 2×(−3)=−62 \times (-3) = -6 and add to 55. These integers are 66 and −1-1. Rewrite the middle term and factor by grouping: 2x2+6x−x−3=2x(x+3)−1(x+3)=(2x−1)(x+3)=02x^2 + 6x - x - 3 = 2x(x + 3) - 1(x + 3) = (2x - 1)(x + 3) = 0.
The equation becomes (2x−1)(x+3)=0(2x - 1)(x + 3) = 0.
Factoring allows us to apply the zero product property to find the solutions.
3
Set each factor of (2x−1)(x+3)=0(2x - 1)(x + 3) = 0 to zero and solve for xx: 2x−1=0⇒x=122x - 1 = 0 \Rightarrow x = \frac{1}{2} and x+3=0⇒x=−3x + 3 = 0 \Rightarrow x = -3.
The solutions are x=12x = \frac{1}{2} and x=−3x = -3, forming the solution set {−3,12}\left\{-3, \frac{1}{2}\right\}.
By the zero product property, if the product of two factors is zero, at least one of the factors must be zero.
4
For the equation 3x2−10x+8=03x^2 - 10x + 8 = 0, factor the trinomial by finding two integers that multiply to 3×8=243 \times 8 = 24 and add to −10-10. These integers are −6-6 and −4-4. Rewrite the middle term and factor by grouping: 3x2−6x−4x+8=3x(x−2)−4(x−2)=(3x−4)(x−2)=03x^2 - 6x - 4x + 8 = 3x(x - 2) - 4(x - 2) = (3x - 4)(x - 2) = 0.
The equation becomes (3x−4)(x−2)=0(3x - 4)(x - 2) = 0.
The equation is already in standard form, so we can directly proceed with factoring.
5
Set each factor of (3x−4)(x−2)=0(3x - 4)(x - 2) = 0 to zero and solve for xx: 3x−4=0⇒x=433x - 4 = 0 \Rightarrow x = \frac{4}{3} and x−2=0⇒x=2x - 2 = 0 \Rightarrow x = 2.
The solutions are x=43x = \frac{4}{3} and x=2x = 2, forming the solution set {43,2}\left\{\frac{4}{3}, 2\right\}.
Solving each linear factor yields the roots of the quadratic equation.
6
For the equation x(x+2)=15x(x + 2) = 15, first distribute xx to get x2+2x=15x^2 + 2x = 15, then subtract 15 from both sides to write in standard form: x2+2x−15=0x^2 + 2x - 15 = 0.
The equation is rewritten as x2+2x−15=0x^2 + 2x - 15 = 0.
Distributing and moving terms sets the quadratic to zero, which is necessary for factoring.
7
Factor the quadratic x2+2x−15=0x^2 + 2x - 15 = 0 by finding two integers that multiply to −15-15 and add to 22. These integers are 55 and −3-3, yielding (x+5)(x−3)=0(x + 5)(x - 3) = 0.
The equation becomes (x+5)(x−3)=0(x + 5)(x - 3) = 0.
Factoring a quadratic trinomial with a leading coefficient of 1 involves finding numbers that sum to the linear coefficient and multiply to the constant term.
8
Set each factor of (x+5)(x−3)=0(x + 5)(x - 3) = 0 to zero and solve for xx: x+5=0⇒x=−5x + 5 = 0 \Rightarrow x = -5 and x−3=0⇒x=3x - 3 = 0 \Rightarrow x = 3.
The solutions are x=−5x = -5 and x=3x = 3, forming the solution set {−5,3}\left\{-5, 3\right\}.
Solving the resulting linear equations gives the roots of the original quadratic equation.

Key Concept

Solving Quadratic Equations by Factoring

Alternative Method

Instead of factoring, the solutions to these quadratic equations can be verified by substituting the values in the solution sets back into the original equations, or by applying the quadratic formula: x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} after rewriting them in standard form.
Estimated Time:2m 0s
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Topics

Solving Quadratic Equations by Factoring

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