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

Question

3888 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-07-19T13:10:45

Separate out the common scalar factor, then spot a perfect square trinomial to find:

$2 x^{2} - 28 x + 98 = 2 (x^{2} - 14 x + 49) = 2 {(x - 7)}^{2}$ $2x^2-28x+98 = 2(x^2-14x+49) = 2(x-7)^2$

First separate out the common scalar factor $2$ $2$ to get:

$2 x^{2} - 28 x + 98 = 2 (x^{2} - 14 x + 49)$ $2x^2-28x+98 = 2(x^2-14x+49)$

Then note that $x^{2} - 14 x + 49$ $x^2-14x+49$ is a perfect square trinomial:

It is of the form $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ $a^2-2ab+b^2 = (a-b)^2$ with $a = x$ $a=x$ and $b = 7$ $b=7$

So $x^{2} - 14 x + 49 = x^{2} - (2 \cdot x \cdot 7) + 7^{2} = {(x - 7)}^{2}$ $x^2-14x+49 = x^2-(2*x*7)+7^2 = (x-7)^2$

How do you factor completely: 2x^2 − 28x + 982x2−28x+982x^2 − 28x + 98?