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

1 Answer

Answer:

#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...)