Question

Evaluate `(2x-3)^5`?

Answer 1

`32x^5-240x^4+720x^3-1080x^2+810x-243`

Explanation:

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

`(2x-3)^5=(2x-3)^2(2x-3)^2(2x-3)`

Let's first work out `(2x-3)^2` using FOIL:

FOIL

• `color(red)(F)` - First terms - `(color(red)(a)+b)(color(red)(c)+d)`
• `color(brown)(O)` - Outside terms - `(color(brown)(a)+b)(c+color(brown)d)`
• `color(blue)(I)` - Inside terms - `(a+color(blue)b)(color(blue)(c)+d)`
• `color(green)(L)` - Last terms - `(a+color(green)b)(c+color(green)d)`

This gives us:

• `color(red)(F)=>(2x)(2x)=4x^2`
• `color(brown)(O)=>(2x)(-3)=-6x`
• `color(blue)(I)=>(-3)(2x)=-6x`
• `color(green)(L)=>(-3)(-3)=9`

`4x^2-6x-6x+9=4x^2-12x+9`

Which means we know have:

`(2x-3)^5=(2x-3)^2(2x-3)^2(2x-3)=(4x^2-12x+9)(4x^2-12x+9)(2x-3)`

Let's go ahead and multiply the two trinomials:

`(4x^2-12x+9)(4x^2-12x+9)=>`

• `(4x^2)(4x^2)=16x^4`
• `(4x^2)(-12x)=-48x^3`
• `(4x^2)(9)=36x^2`
• `(-12x)(4x^2)=-48x^3`
• `(-12x)(-12x)=144x^2`
• `(-12x)(9)=-108x`
• `(9)(4x^2)=36x^2`
• `(9)(-12x)=-108x`
• `(9)(9)=81`

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

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

`(16x^4-96x^3+216x^2-216x+81)(2x-3)=>`

• `(16x^4)(2x)=32x^5`
• `(16x^4)(-3)=-48x^4`
• `(-96x^3)(2x)=-192x^4`
• `(-96x^3)(-3)=288x^3`
• `(216x^2)(2x)=432x^3`
• `(216x^2)(-3)=-648x^2`
• `(-216x)(2x)=-432x^2`
• `(-216x)(-3)=648x`
• `(81)(2x)=162x`
• `(81)(-3)=-243`

`32x^5-48x^4-192x^4+288x^3+432x^3-648x^2-432x^2+648x+162x-243=>`

`32x^5-240x^4+720x^3-1080x^2+810x-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 `(a+b)^n` can be listed as:

`(a+b)^n=C_(n,0)a^nb^0+C_(n,1)a^(n-1)b^1+...+C_(n,n)a^0b^n`

which will give us:

`(2x-3)^5=>`

• `+(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`

`32x^5-240x^4+720x^3-1080x^2+810x-243`

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