How do you factor -x ^ { 2} + 10x + 7?

2 Answers
Jul 5, 2018

-x^2+10x+7=(-x+5+4sqrt(2))(x-5+4sqrt(2))

Explanation:

This does not factor with integer coefficients: consider the quadratic determinant, b^2-4ac, which is equal to 128, which is not a perfect square.

Solve the expression equal to zero to find the non-integer coefficients:
-x^2+10x+7=0
Quadratic formula:
x=-(1/2)(-10+-sqrt(128))=5+-sqrt(32)=5+-4sqrt(2)

So
-x^2+10x+7=(-x+5+4sqrt(2))(x-5+4sqrt(2))

Jul 5, 2018

(4sqrt2-x+5)(4sqrt2+x-5)

Explanation:

"using the method of "color(blue)"completing the square"

=-(x^2-10x-7)

=-(x^2+2(-5)x color(red)(+25)color(red)(-25)-7)

=-(x-5)^2+32

=32-(x-5)^2

"factor using "color(blue)"difference of squares"

•color(white)(x)a^2-b^2=(a-b)(a+b)

"with "a=4sqrt2" and "b=x-5

=(4sqrt2-(x-5))(4sqrt2+x-5)

=(4sqrt2-x+5)(4sqrt2+x-5)