Chapter 6.2, Problem 11MRE

To determine

To calculate: The factors of the trinomial $n^2+5n-14$.

Answer to Problem 11MRE

The trinomial $n^2+5n-14$ is factored as $\left(n+7\right)\left(n-2\right)$.

Explanation of Solution

Given information:

The trinomial $n^2+5n-14$.

Formula used:

To factor a polynomial split the middle term in such a way that sum of two numbers is the middle term and product of two numbers is same is product of first and last term.

Calculation:

Consider the trinomial $n^2+5n-14$.

Recall that to factor a polynomial split the middle term in such a way that sum of two numbers is the middle term and product of two numbers is same is product of first and last term.

For the above polynomial $7$ and $-2$ are two numbers such that their sum is $5$ and product is $-14$.

Therefore, the trinomial is rewritten as,

  $n^2+5n-14=n^2+7n-2n-14$

Take out the common terms,

  $\begin{array}{c}{n}^2+5n-14=n^2+7n-2n-14\\ =n\left(n+7\right)-2\left(n+7\right)\\ =\left(n+7\right)\left(n-2\right)\end{array}$

Thus, thetrinomial $n^2+5n-14$ is factored as $\left(n+7\right)\left(n-2\right)$.