Q. 356 - Intermediate Algebra

Question

In the following exercises, factor each trinomial of the form $x^2+bx+c$

$k^2-16k+60$

Step-by-Step Solution

Answer

The factored form of the polynomial is,

$\left(k-10\right)\left(k-6\right)$

Step 1. Given Information

We are given a polynomial,  

$k^2-16k+60$

Step 2. Factorizing the polynomial

Using the product of $ac$ method in factorizing the polynomial of form $a{x}^2+bx+c$, we get

$$\begin{gathered} 10\times 6=60 \\ 10+6=16 \end{gathered}$$

So, Splitting the middle term,

$k^2-16k+60=k^2-10k-6k+60$

Grouping terms, we get 

$k^2-16k+60=\left(k^2-10k\right)-\left(6k-60\right)$

Factoring out GCF from groups, we get 

$k^2-16k+60=k\left(k-10\right)-6\left(k-10\right)$

Again, factoring out GCF, we get 

$k^2-16k+60=\left(k-10\right)\left(k-6\right)$

Step 3. Checking the factorization

 Multiplying the factors, 

$$\begin{gathered} \left(k-10\right)\left(k-6\right)=k^2-16k+60 \\ k\left(k-6\right)-10\left(k-6\right)=k^2-16k+60 \\ {k}^2-6k-10k+60=k^2-16k+60 \\ {k}^2-16k+60=k^2-16k+60 \\ LHS=RHS \end{gathered}$$

Hence the factorization is correct.