Q. 387 - Intermediate Algebra

Question

In the following exercise, factor completely using the perfect square trinomials pattern.   
$75u^4-30u^{3}v+3u^2{v}^2$

Step-by-Step Solution

Answer

The factorisation of the given polynomial is:

$75u^4-30u^{3}v+3u^2{v}^2=3u^2{(5u-v)}^2$

Step 1. Given information

The given polynomial is:  

$75u^4-30u^{3}v+3u^2{v}^2$

Step 2. Factorise the polynomial

We know that,  

${\left(a+b\right)}^2=a^2-2ab+b^2$

For the polynomial $75u^4-30u^{3}v+3u^2{v}^2$, taking out $3u^2$ as a common, we get:

$3u^2(25u^2-10uv+3v^2)$

Now,

$$\begin{gathered} 3u^2(25u^2-10uv+v^2) \\ 3u^2[{\left(5u\right)}^2-2\times 5u\times v+v^2] \\ 3u^2{(5u-v)}^2 [u\sin g {(a-b)}^2=a^2-2ab+b^2] \end{gathered}$$

The factorisation of the given polynomial is:

$75u^4-30u^{3}v+3u^2{v}^2=3u^2{(5u-v)}^2$

Step 3. Check the answer

Multiplying the factors we get:

$$\begin{gathered} 75u^4-30u^{3}v+3u^2{v}^2=3u^2{(5u-v)}^2 \\ 75u^4-30u^{3}v+3u^2{v}^2=3u^2(5u-v)(5u-v) \\ 75u^4-30u^{3}v+3u^2{v}^2=3u^2(25u^2-10uv+v^2) \\ 75u^4-30u^{3}v+3u^2{v}^2=75u^4-30u^{3}v+3u^2{v}^2 \end{gathered}$$

This is true.