How do you factor 100+4x^2-16y-40x100+4x2−16y−40x?
1 Answer
If the
100+4x^2-16y^2-40x = 4(x-2y-5)(x+2y-5)100+4x2−16y2−40x=4(x−2y−5)(x+2y−5)
Explanation:
For the record: I think the question should have specified
The difference of squares identity can be written:
a^2-b^2 = (a-b)(a+b)a2−b2=(a−b)(a+b)
Use this with
100+4x^2-16y^2-40x = 4(x^2-10x+25-4y^2)100+4x2−16y2−40x=4(x2−10x+25−4y2)
color(white)(100+4x^2-16y^2-40x) = 4(x^2-2x(5)+5^2-4y^2)100+4x2−16y2−40x=4(x2−2x(5)+52−4y2)
color(white)(100+4x^2-16y^2-40x) = 4((x-5)^2-(2y)^2)100+4x2−16y2−40x=4((x−5)2−(2y)2)
color(white)(100+4x^2-16y^2-40x) = 4((x-5)-2y)((x-5)+2y)100+4x2−16y2−40x=4((x−5)−2y)((x−5)+2y)
color(white)(100+4x^2-16y^2-40x) = 4(x-2y-5)(x+2y-5)100+4x2−16y2−40x=4(x−2y−5)(x+2y−5)