Exercise 70 Page 287

To factor a trinomial with a leading coefficient of $1,$ think of the process as multiplying two binomials in reverse. Let's start by taking a look at the constant term. In this case, we have $\text{-}20.$ This is a negative number, so for the product of the constant terms in the factors to be negative, these constants must have the opposite sign (one positive and one negative).

Factor Constants	Product of Constants
$1$ and $\text{-}20$	$\text{-}20$
$\text{-}1$ and $20$	$\text{-}20$
$2$ and $\text{-}10$	$\text{-}20$
$\text{-}2$ and $10$	$\text{-}20$
$4$ and $\text{-}5$	$\text{-}20$
$\text{-}4$ and $5$	$\text{-}20$

Next, let's consider the coefficient of the linear term.

$$\begin{array}{c}{x}^2+8x-20\end{array}$$

For this term, we need the sum of the factors that produced the constant term to equal the coefficient of the linear term, $8.$

Factors	Sum of Factors
$1$ and $\text{-}20$	$\text{-}19$
$\text{-}1$ and $20$	$19$
$2$ and $\text{-}10$	$\text{-}8$
$\text{-}2$ and $10$	$8$
$4$ and $\text{-}5$	$\text{-}1$
$\text{-}4$ and $5$	$1$

We found the factors whose product is $20$ and whose sum is $8.$

$$\begin{array}{c}{x}^2+8x-20\iff (x-2)(x+10)\end{array}$$

Checking Our Answer

Check your answer $✓$

We can check our answer by applying the Distributive Property and comparing the result with the given expression.

$(x-2)(x+10)$

Distr

Distribute $(x+10)$

$x\left(x+10\right)-2\left(x+10\right)$

Distr

Distribute $x$

$x^2+10x-2\left(x+10\right)$

Distr

Distribute $(\text{-}2)$

$x^2+10x-2x-20$

SubTerm

Subtract term

$x^2+8x-20$

After applying the Distributive Property and simplifying, the result is the same as the given expression. Therefore, we can be sure our solution is correct!