algebra II: x^2 - 7xy-8y^2 (factor)? thanks for any help....?

x^2 - 7xy-8y^2
=(x+y)(x-8y) answer//

Write 2 gadgets of parentheses. (_________)(_________) = look on the 1st term's factors. y^6 = y^0 * y^6 = y * y^5 = y^2 * y^4 = y^3 * y^3 = y^4 * y^2 = y^5 * y = y^6 * y^0 because of the fact the middle term is y^3, all of us comprehend that we want y^3 * y^3 because of the fact the factors. placed y^3 on the initiating of each set of parentheses. (y^3_____)(y^3_____) = look on the final term's factors. you prefer the pair which will upload as much as the middle coefficient: 7. Multiples a million*-8 ==> a million + -8 = -7 <== incorrect pair 2*-4 ==> 2 + -4 = -2 <== incorrect pair 4*-2 ==> 4 + -2 = 2 <== incorrect pair 8*-a million ==> 8 + -a million = 7 <== good pair Write 8 and -a million interior the the rest blanks. (y^3 + 8)(y^3 - a million) = word that y^3 + 8 is the sum of two squares. y^3 + 8 = (y)^3 + (2)^3 all of us comprehend that a^3 + b^3 = (a + b)(a^2 - ab + b^2) Given: y^3 + 8 = (y)^3 + (2)^3 skill: a = y, b = 2 replace y^3 + 8 with the factored type. (y^3 + 8)(y^3 - a million) = (y + 2)(y^2 - 2y + 4)(y^3 - a million) = word that y^3 - a million is the version of two squares. y^3 - a million = (y)^3 - (a million)^3 all of us comprehend that a^3 - b^3 = (a - b)(a^2 + ab + b^2) Given: y^3 - a million = (y)^3 - (a million)^3 skill: a = y, b = a million replace y^3 - a million with the factored type. (y + 2)(y^2 - 2y + 4)(y^3 - a million) = (y + 2)(y^2 - 2y + 4)(y - a million)(y^2 + y + a million) answer: (y + 2)(y^2 - 2y + 4)(y - a million)(y^2 + y + a million)

(x+y)(x-8y)

x^2 - 7xy - 8y^2
= x^2 + xy - 8xy - 8y^2
= (x^2 + xy) - (8xy + 8y^2)
= x(x + y) - 8y(x + y)
= (x + y)(x - 8y)

x^2 - 7xy - 8y^2
=x^2 - 8xy + xy - 8y^2
=x(x - 8y) + y(x - 8y)
=(x - 8y)(x + y)

=(x-8y)(x+y)
God bless you.

(x - 8y)(x + y)!

x^2 - 7xy-8y^2
= x^2 - 8xy + xy - 8y^2
= x(x - 8y) + y( x - 8y)
= (x + y)(x - 8y) (Answer

The answer would be (x-8y)(x+y)