Question

Solve by Factoring

1. x^2-x-20=0

2. x(14x-15)=9

Answer 1

Solve by Factoring

1. $x^{2} - x - 20 = 0$

2. $x(14x - 15) = 9$

**Solution:**

A quadratic equation is a polynomial that looks like $ax^2 + bx + c$, where $a, b,$ and $c$ are numbers. For the easy case of factoring, we will find two numbers that will not only multiply to equal the constant term $c$, but also add up to equal $b$, the coefficient on the x-term.

This equation is already in the form quadratic equals zero: $x^{2} - x - 20 = 0$.

We need to find factors of $-20$ that add up to 1. Since $-20$ can be written as the product of -4 and 5, and since $5 + (-4) = 1$, then we will use -4 and 5. From multiplying polynomials that this quadratic is formed from multiplying two factors of the form $(x + m)(x + n)$, for some numbers $m$ and $n$. We will write in the two numbers that we found above:

$$(x + 4)(x - 5) = 0$$

Solve each factor: $(x + 4) = 0, x = -4; (x - 5) = 0, x = 5$.

The solutions to $x^{2} - x - 20 = 0$ are $x = 5, -4$

Checking $x = 5$ in $x^{2} - x - 20 = 0$

$$\begin{array}{l} (5)^{2} - 5 - 20 = 0 \\ 25 - 5 - 20 = 0 \\ 20 - 20 = 0 \\ 0 = 0 \\ \end{array}$$

Checking $x = -4$ in $x^{2} - x - 20 = 0$

$$\begin{array}{l} (-4)^{2} - (-4) - 20 = 0 \\ 16 + 4 - 20 = 0 \\ 20 - 20 = 0 \\ 0 = 0 \\ \end{array}$$

We check both solutions of quadratic equation.

**Another variant to solve equation**

To solve this problem we multiplying $a$ and $c$ ($a = 1, c = -20$). We get $(1)(-20) = -20$.

We can subtract the pairs to find the differences. If there is a pair of factors with a difference of 1, then we can factor the quadratic. Now that we have factor pair (with the larger number having the "minus" sign), factor the quadratic:

Subtract into quadratic equation $x^{2} - x - 20 = x^{2} + 4x - 5x - 20 = x(x + 4) - 5(x + 4) = (x - 5)(x + 4) = 0$

The solutions to $x^{2} - x - 20 = 0$ are $x = 5, -4$ .

2. $x(14x - 15) = 9$

Compare our equation to the standard form, $ax^2 + bx + c$ : $14x^2 - 15x - 9 = 0$

Identify values $a, b$ and $c$ . ( $a = 14, b = -15$ and $c = -9$ ). Find a pair of factors of $a \times c = 14 \times (-9) = -126$ , whose sum is $b = -15$ .

We note the possibilities in a table:

Using the factor pairs [6, -21], which get sum $b$ equal $= -15$ . Rewrite our equation replacing the term $-15x$ with $6x$ and $-21x$ .

Subtract into equation $14x^{2} - 15x - 9 = 14x^{2} + 6x - 21x - 9 = 0$

$$1 4 x ^ {2} + 6 x - 2 1 x - 9 = 0$$

Group the first two terms and the last two terms on the left side:

$$(14x^2 + 6x) + (-21x - 9) = 0$$

Factor common factors:

$$2x(7x + 3) - 3(7x + 3) = 0$$

The two quantities in parentheses are the same. We have common quantity $(7x + 3)$, so we can factor it out:

$$(7x + 3)(2x - 3) = 0$$

Solve each factor: $(7x + 3) = 0, 7x = -3, x = -\frac{3}{7}; (2x - 3) = 0, 2x = 3, x = \frac{3}{2}$.

The solutions to $x^2 - x - 20 = 0$ are $x = -\frac{3}{7}, \frac{3}{2}$

Checking $x = -\frac{3}{7}$ in $14x^2 - 15x - 9 = 0$

$$14 \cdot \left(-\frac{3}{7}\right)^2 - 15 \cdot \left(-\frac{3}{7}\right) - 9 = 0$$ $$\left(14 \cdot \frac{9}{49}\right) + \frac{45}{7} - 9 = 0$$ $$\frac{18}{7} + \frac{45}{7} - 9 = 0$$ $$\frac{63}{7} - 9 = 0$$ $$9 - 9 = 0$$

Checking $x = \frac{3}{2}$ in $14x^2 - 15x - 9 = 0$

$$14 \cdot \left(\frac{3}{2}\right)^2 - 15 \cdot \frac{3}{2} - 9 = 0$$ $$14 \cdot \frac{9}{4} - \frac{45}{2} - 9 = 0$$ $$\frac{63}{2} - \frac{45}{2} - 9 = 0$$ $$\frac{18}{2} - 9 = 0$$ $$9 - 9 = 0$$

The solutions to $14x^2 - 15x - 9 = 0$ are $x = \frac{3}{2}, x = -\frac{3}{7}$.