Question

How do you factor completely: $2{{x}^{2}}-28x+98$?

Answer 1

Hint: We will look at the concept of factoring a polynomial. We will simplify the given polynomial by taking out common factors. Then we will split the middle term in such a manner that there will be a common factor in two terms and another common factor in the other two terms. We will take out these common terms and obtain the factors of the given polynomial.

Complete step by step answer:
The given polynomial is $2{{x}^{2}}-28x+98$. Now, we can see that all the three terms are divisible by 2. So, we will take out 2 as a common factor. We obtain the following expression,
$2{{x}^{2}}-28x+98=2\left( {{x}^{2}}-14x+49 \right)$
Now, the middle term is $-14$. We can split the middle term as $-14x=-7x-7x$. So, the expression we have will become
$2{{x}^{2}}-28x+98=2\left( {{x}^{2}}-7x-7x+49 \right)$
In the above expression, on the right hand side, we can see that there are pairs of terms that have a common factor. The first and the second term have $x$ as a common factor. Since $49=-7\times -7$ we can see that the third and the fourth term have $-7$ as a common factor. Therefore, we can write the right hand side of the above expression as the following,
$2{{x}^{2}}-28x+98=2\left( x\left( x-7 \right)-7\left( x-7 \right) \right)$
Now, from inside the outer bracket, we can take $\left( x-7 \right)$ as a common factor.

Therefore we get the following,
$\begin{align}
  & 2{{x}^{2}}-28x+98=2\left( x-7 \right)\left( x-7 \right) \\
 & \therefore 2{{x}^{2}}-28x+98=2{{\left( x-7 \right)}^{2}} \\
\end{align}$

Note: We can use the algebraic identity ${{\left( x+y \right)}^{2}}={{x}^{2}}+2xy+{{y}^{2}}$ to obtain the factors of the given polynomial. After taking out 2 as the common factor, we are left with the polynomial, ${{x}^{2}}-14x+49$. We can see that ${{x}^{2}}-14x+49={{x}^{2}}-2\times 7\times x+{{7}^{2}}$. So, comparing it with the identity mentioned before, we can see that the expression is clearly the expansion of ${{\left( x-7 \right)}^{2}}$. We arrive at the same solution using this method as well.