Evaluate (2x-3)^5?

$32 {x}^{5} - 240 {x}^{4} + 720 {x}^{3} - 1080 {x}^{2} + 810 x - 243$

Explanation:

Let's break this down a bit by rewriting the expression:

${\left(2 x - 3\right)}^{5} = {\left(2 x - 3\right)}^{2} {\left(2 x - 3\right)}^{2} \left(2 x - 3\right)$

Let's first work out ${\left(2 x - 3\right)}^{2}$ using FOIL:

FOIL

• $\textcolor{red}{F}$ - First terms - $\left(\textcolor{red}{a} + b\right) \left(\textcolor{red}{c} + d\right)$
• $\textcolor{b r o w n}{O}$ - Outside terms - $\left(\textcolor{b r o w n}{a} + b\right) \left(c + \textcolor{b r o w n}{d}\right)$
• $\textcolor{b l u e}{I}$ - Inside terms - $\left(a + \textcolor{b l u e}{b}\right) \left(\textcolor{b l u e}{c} + d\right)$
• $\textcolor{g r e e n}{L}$ - Last terms - $\left(a + \textcolor{g r e e n}{b}\right) \left(c + \textcolor{g r e e n}{d}\right)$

This gives us:

• $\textcolor{red}{F} \implies \left(2 x\right) \left(2 x\right) = 4 {x}^{2}$
• $\textcolor{b r o w n}{O} \implies \left(2 x\right) \left(- 3\right) = - 6 x$
• $\textcolor{b l u e}{I} \implies \left(- 3\right) \left(2 x\right) = - 6 x$
• $\textcolor{g r e e n}{L} \implies \left(- 3\right) \left(- 3\right) = 9$

$4 {x}^{2} - 6 x - 6 x + 9 = 4 {x}^{2} - 12 x + 9$

Which means we know have:

${\left(2 x - 3\right)}^{5} = {\left(2 x - 3\right)}^{2} {\left(2 x - 3\right)}^{2} \left(2 x - 3\right) = \left(4 {x}^{2} - 12 x + 9\right) \left(4 {x}^{2} - 12 x + 9\right) \left(2 x - 3\right)$

Let's go ahead and multiply the two trinomials:

$\left(4 {x}^{2} - 12 x + 9\right) \left(4 {x}^{2} - 12 x + 9\right) \implies$

• $\left(4 {x}^{2}\right) \left(4 {x}^{2}\right) = 16 {x}^{4}$
• $\left(4 {x}^{2}\right) \left(- 12 x\right) = - 48 {x}^{3}$
• $\left(4 {x}^{2}\right) \left(9\right) = 36 {x}^{2}$
• $\left(- 12 x\right) \left(4 {x}^{2}\right) = - 48 {x}^{3}$
• $\left(- 12 x\right) \left(- 12 x\right) = 144 {x}^{2}$
• $\left(- 12 x\right) \left(9\right) = - 108 x$
• $\left(9\right) \left(4 {x}^{2}\right) = 36 {x}^{2}$
• $\left(9\right) \left(- 12 x\right) = - 108 x$
• $\left(9\right) \left(9\right) = 81$

$16 {x}^{4} - 48 {x}^{3} + 36 {x}^{2} - 48 {x}^{3} + 144 {x}^{2} - 108 x + 36 {x}^{2} - 108 x + 81 = 16 {x}^{4} - 96 {x}^{3} + 216 {x}^{2} - 216 x + 81$

And now to complete things, let's multiply the last of the brackets:

$\left(16 {x}^{4} - 96 {x}^{3} + 216 {x}^{2} - 216 x + 81\right) \left(2 x - 3\right) \implies$

• $\left(16 {x}^{4}\right) \left(2 x\right) = 32 {x}^{5}$
• $\left(16 {x}^{4}\right) \left(- 3\right) = - 48 {x}^{4}$
• $\left(- 96 {x}^{3}\right) \left(2 x\right) = - 192 {x}^{4}$
• $\left(- 96 {x}^{3}\right) \left(- 3\right) = 288 {x}^{3}$
• $\left(216 {x}^{2}\right) \left(2 x\right) = 432 {x}^{3}$
• $\left(216 {x}^{2}\right) \left(- 3\right) = - 648 {x}^{2}$
• $\left(- 216 x\right) \left(2 x\right) = - 432 {x}^{2}$
• $\left(- 216 x\right) \left(- 3\right) = 648 x$
• $\left(81\right) \left(2 x\right) = 162 x$
• $\left(81\right) \left(- 3\right) = - 243$

$32 {x}^{5} - 48 {x}^{4} - 192 {x}^{4} + 288 {x}^{3} + 432 {x}^{3} - 648 {x}^{2} - 432 {x}^{2} + 648 x + 162 x - 243 \implies$

$32 {x}^{5} - 240 {x}^{4} + 720 {x}^{3} - 1080 {x}^{2} + 810 x - 243$

~~~~~~~~~~

This was quite the undertaking - I'm going to check my work by using a different method - the Binomial Theorem, which states that the terms in an expansion of the form ${\left(a + b\right)}^{n}$ can be listed as:

${\left(a + b\right)}^{n} = {C}_{n , 0} {a}^{n} {b}^{0} + {C}_{n , 1} {a}^{n - 1} {b}^{1} + \ldots + {C}_{n , n} {a}^{0} {b}^{n}$

which will give us:

${\left(2 x - 3\right)}^{5} \implies$

• +(5!)/((0!)(5!))(2x)^5(-3)^0=1(32x^5)(1)=32x^5
• +(5!)/((1!)(4!))(2x)^4(-3)^1=5(16x^4)(-3)=-240x^4
• +(5!)/((2!)(3!))(2x)^3(-3)^2=10(8x^3)(9)=720x^3
• +(5!)/((3!)(2!))(2x)^2(-3)^3=10(4x^2)(-27)=-1080x^2
• +(5!)/((4!)(1!))(2x)^1(-3)^4=5(2x)(81)=810x
• +(5!)/((5!)(0!))(2x)^0(-3)^5=(1)(1)(-243)=-243

$32 {x}^{5} - 240 {x}^{4} + 720 {x}^{3} - 1080 {x}^{2} + 810 x - 243$

and it's from here that I can see I made a mistake above! (which I've now edited out...)