Chapter SH, Problem 14.8.6EP

To determine

To factor: the given polynomial

Answer to Problem 14.8.6EP

The required result is $\left(x-3\right)\left(x+9\right)$ .

Explanation of Solution

Given: $x^2+6x-27$

Calculation:

Write the trinomial as the product of two binomials.

Product of first terms is $x^2$ .

  $x^2+6x-27=\left(x+?\right)\left(x+?\right)$

Now find the two terms whose product must be $-27$ .

Try several factors of $-27$ so that sum of the product of outer and inner terms must be $6x$ .

Try the factors $3,-9$ .

  $\begin{array}{l}\left(x+3\right)\left(x-9\right)=x^2+3x-9x-27\\ =x^2-6x-27\end{array}$

It is not true.

Try the factor $3,-9$ .

  $\begin{array}{l}\left(x-3\right)\left(x+9\right)=x^2-3x+9x-27\\ =x^2+6x-27\end{array}$

It is true.

Hence, the result is $\left(x-3\right)\left(x+9\right)$ .

Conclusion:

Hence, the factorization is $\left(x-3\right)\left(x+9\right)$ .