How do you factor completely #2x^4 -14x^2 +24#?

Question

Abigail Allen · Accepted Answer

#color(green)(2(x-sqrt(3))(x+sqrt(3))(x-2)(x+2))# Temporally replacing #x^2# with #a# #2x^4-14x^2+24# becomes #2a^2-14a+24# #2a^2-14a+24# #color(white)("XXX")=2(a^2-7a+12)# #color(white)("XXX")=2(a-3)(a-4)# Restoring #x^2# in place of #a# #color(white)("XXX"){: ((a-3)=(x^2-3),color(white)("XX"),(a-4)=(x^2-4)), (color(white)("X")=(x-sqrt(3))(x+sqrt(3)),,color(white)("X")=(x-2)(x+2)) :}# So #2x^4-14x^2+24# #color(white)("XXX")=color(green)(2(x-sqrt(3))(x+sqrt(3))(x-2)(x+2))#

Eva Adamson · Answer

undefined To factor completely the expression $2x^4 - 14x^2 + 24$, we first look for a common factor. In this case, the greatest common factor (GCF) is 2. Factoring out 2 gives:

$$ 2(x^4 - 7x^2 + 12) $$

Now, we need to factor the quadratic inside the parentheses. To do this, we look for two numbers that multiply to 12 (the constant term) and add up to -7 (the coefficient of the $x^2$ term). These numbers are -4 and -3. So, we can rewrite the expression as:

$$ 2(x^4 - 4x^2 - 3x^2 + 12) $$

Now, we factor by grouping:

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

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

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

Therefore, the completely factored form of $2x^4 - 14x^2 + 24$ is $2(x^2 - 3)(x^2 - 4)$.