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

2 Answers
MoominDave
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))#

Jim G.
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)#

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