How do you factor #2x^5+4x^4-4x^3-8x^2# ?

Question

Nathan Axel · Accepted Answer

#2x^5+4x^4-4x^3-8x^2 = 2x^2(x-sqrt(2))(x+sqrt(2))(x+2)#

Notice that the ratio between the first and second terms is the same as that between the third and fourth terms. So this quadrinomial will factor by grouping.

Note that all of the terms are divisible by #2x^2# so that can be separated out first.

Finally we can use the difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

with #a=x# and #b=sqrt(2)#

#2x^5+4x^4-4x^3-8x^2 = 2x^2(x^3+2x^2-2x-4)#

#color(white)(2x^5+4x^4-4x^3-8x^2) = 2x^2((x^3+2x^2)-(2x+4))#

#color(white)(2x^5+4x^4-4x^3-8x^2) = 2x^2(x^2(x+2)-2(x+2))#

#color(white)(2x^5+4x^4-4x^3-8x^2) = 2x^2(x^2-2)(x+2)#

#color(white)(2x^5+4x^4-4x^3-8x^2) = 2x^2(x^2-(sqrt(2))^2)(x+2)#

#color(white)(2x^5+4x^4-4x^3-8x^2) = 2x^2(x-sqrt(2))(x+sqrt(2))(x+2)#

Julia Adams · Answer

undefined To factor the expression 2x^5 + 4x^4 - 4x^3 - 8x^2, you can start by factoring out the greatest common factor, which is 2x^2. After factoring out 2x^2, you will have:

2x^2(x^3 + 2x^2 - 2x - 4)

Next, you can factor the expression inside the parentheses using various methods such as grouping, synthetic division, or trial and error.