Question

In the following exercise, factor completely using the perfect square trinomials pattern.

$5k^3-70k^2+245k$

Short answer

The factorisation of the polynomial is

$5k^3-70k^2+245k=5k{(k-7)}^2$

Step-by-step solution

Step 1. Given information 

The given polynomial is:

$5k^2-70k^2+245k$

Step 2. Factorise the polynomial  

We know that,

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

For the polynomial $5k^3-70k^2+245k$, taking out $5k$as a common, we get:

$5k(k^2-14k+49)$

Now,

localid="1644658099435" $$\begin{gathered} 5k[k^2-14k+49] \\ 5k[{\left(k\right)}^2-2\times k\times 7+7^2] \\ 5k{(k-7)}^2[u\sin g{(a-b)}^2=a^2-2ab+b^2] \end{gathered}$$

Thus, the factorisation of the polynomial is:

localid="1644657063485" $5k^3-70k^2+245k=5k{(k-7)}^2$

Step 3. check the answer

Multiplying the factors, we get:

$$\begin{gathered} 5k^3-70k^2+245k=5k{(k-7)}^2 \\ 5k^3-70k^2+245k=5k(k^2-14k+49) \\ 5k^3-70k^2+245k=5k^3-70k^2+245k \end{gathered}$$

This is true.