Factor each of the following expressions completely and then simplify, if possible:
a: ax + 2x
b: ax - 2x
c: 3x - 4x + 7x
d: 3x^2 + xy - x
e: (a + b)(c + 1) - (a + b)
f: x^2 - 9y^2
g: (x^2 - y^2) + x + y
4 answers
A: x(a+2)
a: ax+2x = x(a+2)
b: ax-2x = x(a-2)
c: 3x-4x+7x = x(3-4+7) = 6x
d: 3x^2+xy-x = x(3x+y-1)
e: (a+b)(c+1)-(a+b) = ac+bc = c(a+b)
f: x^2-9y^2 = x=3y
g:(x^2 - y^2) + x + y = y-x
b: ax-2x = x(a-2)
c: 3x-4x+7x = x(3-4+7) = 6x
d: 3x^2+xy-x = x(3x+y-1)
e: (a+b)(c+1)-(a+b) = ac+bc = c(a+b)
f: x^2-9y^2 = x=3y
g:(x^2 - y^2) + x + y = y-x
How did you do this??
Niezel is correct from a - e
explanation of e)
(a+b)(c+1)-(a+b) = ac+bc = c(a+b)
by common factor of (a_+b)
= (a+b)(c+1 - 1)
= c(a+b)
f)
x^2 - 9y^2 ----- standard difference of squares
= (x+3y)(x-3y)
g)
(x^2 - y^2) + x + y
= (x-y)(x+y) + (x+y) , difference of squares at the front, no damage done by putting in brackets at the end
= (x+y)(x-y + 1) --- just used a common factor of (x+y)
explanation of e)
(a+b)(c+1)-(a+b) = ac+bc = c(a+b)
by common factor of (a_+b)
= (a+b)(c+1 - 1)
= c(a+b)
f)
x^2 - 9y^2 ----- standard difference of squares
= (x+3y)(x-3y)
g)
(x^2 - y^2) + x + y
= (x-y)(x+y) + (x+y) , difference of squares at the front, no damage done by putting in brackets at the end
= (x+y)(x-y + 1) --- just used a common factor of (x+y)
2 answers