Chapter 8.6, Problem 2BGP

To determine

To factor: The polynomial $w^2-11w+28$

Answer to Problem 2BGP

  $\left(w-7\right)\left(w-4\right)$

Explanation of Solution

Given:

  $w^2-11w+28$

Calculation:

The given polynomial is $w^2-11w+28$

Rewrite the middle term as $-11w=-7w-4w$ . So, the polynomial becomes

  $w^2-7w-4w+28$

Make group as shown below

  $\left(w^2-7w\right)+\left(-4w+28\right)$

Factor out the GCF from each of the group

  $w\left(w-7\right)-4\left(w-7\right)$

Finally, factored out the common factor

  $\left(w-7\right)\left(w-4\right)$

Therefore, the factored form of the given polynomial is $\left(w-7\right)\left(w-4\right)$