Chapter 5.11, Problem 4WE

To determine

To factor the below expression completely:

  $6ab+9a^{2}b-15a^{3}b$

Answer to Problem 4WE

  $3ab\left(1-a\right)\left(2+5a\right)$

Explanation of Solution

Given:

  $6ab+9a^{2}b-15a^{3}b$

Concept Used:

To factor a polynomial completely:

• Factor out the greatest monomial factor first.
• Look for a difference of squares and a perfect square trinomial.
• If a trinomial is not a square, look for a pair of binomial factors.
• Make sure that each binomial or trinomial factor is prime.

Calculation:

To factor the expression:

  $6ab+9a^{2}b-15a^{3}b$

First, factor out the greatest monomial factor that is $3ab$ , as

  $\begin{array}{c}6ab+9a^{2}b-15a^{3}b=3ab\times 2+3ab\times 3a-3ab\times 5a^2\\ =3ab\left(2+3a-5a^2\right)\end{array}$

Now, the second factor, $\left(2+3a-5a^2\right)$ can be factorized further by grouping, as

  $2+3a-5a^2=2-2a+5a-5a^2$

Take common factor 2 from first two terms out, and take common factor $5a$ from last two terms out and simplify further, as

  $\begin{array}{c}2-2a+5a-5a^2=2\left(1-a\right)+5a\left(1-a\right)\\ =\left(1-a\right)\left(2+5a\right)\end{array}$

Now, note that each factors $3ab,\left(1-a\right),\text{ and }\left(2+5a\right)$ are prime, i.e. can’t be factorize further.

Thus, factor of given polynomial is:

  $6ab+9a^{2}b-15a^{3}b=3ab\left(1-a\right)\left(2+5a\right)$