How do you factor #4m^4 - 37m^2 + 9#?

Question

Julia Adams · Accepted Answer

#4m^4-37m^2+9 = (2m-1)(2m+1)(m-3)(m+3)#

The difference of squares identity can be written:

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

We will use this a couple of times, but first note that:

#4m^4-37m^2+9 = 4(m^2)^2-37(m^2)+9#

So we can treat this quartic as a quadratic in #m^2#

Factor using an AC method:

Look for a pair of factors of #AC = 4*9 = 36# with sum #B=37#

The pair #36, 1# works.

Use this pair to split the middle term and factor by grouping:

#4(m^2)^2-37(m^2)+9 = 4(m^2)^2-36(m^2)-(m^2)+9#

#color(white)(4(m^2)^2-37(m^2)+9) = (4(m^2)^2-36(m^2))-((m^2)-9)#

#color(white)(4(m^2)^2-37(m^2)+9) = 4m^2(m^2-9)-1(m^2-9)#

#color(white)(4(m^2)^2-37(m^2)+9) = (4m^2-1)(m^2-9)#

#color(white)(4(m^2)^2-37(m^2)+9) = ((2m)^2-1^2)(m^2-3^2)#

#color(white)(4(m^2)^2-37(m^2)+9) = (2m-1)(2m+1)(m-3)(m+3)#

Abigail Allen · Answer

#(4m^2 - 1)(m^2 - 9)# Factoring this polynomial gives: #(4m^2 - 1)(m^2 - 9)#

Avery Alexander · Answer

undefined To factor $4m^4 - 37m^2 + 9$, first, let $x = m^2$. Then, rewrite the expression as $4x^2 - 37x + 9$. Now, factor $4x^2 - 37x + 9$ as $(4x - 1)(x - 9)$. Substitute back $x = m^2$ to get $(4m^2 - 1)(m^2 - 9)$. Further factor $4m^2 - 1$ as $(2m + 1)(2m - 1)$ and $m^2 - 9$ as $(m + 3)(m - 3)$. Therefore, the factored form is $(2m + 1)(2m - 1)(m + 3)(m - 3)$.