# Solving Quadratic Equations by Factoring

By | December 23, 2015

In the previous post, we have learned how to solve quadratic equations by extracting the roots. In this post, we are going to learn how to solve quadratic equations by factoring.

To solve quadratic equations by factoring, we need to use the zero property of real numbers. It states that the product of two real numbers is zero if at least one of the two real numbers is zero. In effect, we need to transfer all the terms to the lefth hand side, let the right hand side be 0, equate factored form zero and find the value of x.

Example 1: $2x(x + 5) = 0$

Solution
$2x = 0, x = 0$
$x + 5 = 0, x = -5$.

This means the solutions of $2x(x+5) =0$ are $0$ or $-5$.

Example 2: $x^2 - x = 6$

Solution
Subtracting $6$ from both sides, we have
$x^2 - x - 6 = 0$.

Factoring, we have
$(x - 3)(x + 2) = 0$
$x - 3 = 0, x = 3$
$x + 2 = 0, x = -2$.

This means the solutions of $x^2 - x = 6$ are $3$ or $-2$.

Example 3: $x^2 = x \sqrt{3}$

Solution

Subtracting $\sqrt{3}$ from both sides,
$x^2 - x \sqrt{3} = 0$.

Factoring out $x$, we have
$x(x - \sqrt{3}) = 0$.

Equating both expression to 0, we have
$x = 0$
$x - \sqrt{3} = 0, x = \pm \sqrt{3}$.

This means the solutions of $x^2 - x = 6$ are $0$ or $\pm \sqrt{3}$.

Example 4: Solve $x^2 = 16$.

Solution

$x^2 - 16 = 0$
$(x + 4)(x - 4) = 0$
$x = -4, x = 4$.

This means the solutions of $x^2 = 16$ are $-4$ or $4$.

Example 5: Solve $x^2 + 2x + 1 = 0$.

Solution
$x^2 + 2x + 1 = 0$
$(x + 1)(x + 1) = 0$
$x + 1 = 0, x = -1$
$x + 1 = 0, x = -1$
$x = -1, x = -1$.

This means the solutions of $x^2 + 2x + 1 = 0$ are $katex -1$ or $-1$.

Example 6: Solve $(x - 3)^2 = 4x$

Solution
$(x - 3)^2 = 4x$
$x^2 - 6x + 9 = 4x$
$x^2 - 6x - 4x + 9 = 0$
$x^2 - 10x + 9 = 0$
$(x - 1)(x - 9) = 0$
$x = 1, x = 9$.

This means the solutions of $(x - 3)^2 = 4x$ are $1$ or $9$.

Example 7: $\frac{2x - 3}{x + 3} = \frac{x + 3}{x - 3}$
Solution

Cross multiplying, we have
$(2x - 3)(x - 3) = (x + 3)(x + 3)$.

Expanding, we have
$2x^2 - 9x + 9 = x^2 + 6x + 9$.

Transposing all the terms to the left hand side, we have
$x^2 - 15x = 0$
$x(x - 15) = 0$
$x = 0$
$x - 15 = 0$
$x = 15$.

This means the solutions are $0$ or $15$.