1) By equating like terms, a=1 since the coefficients of x3 must be equal. 2) Equating the coefficients of x2 gives -3+b=5, so b=8. 3) Equating the coefficients of x gives -13+c=-8, so c=5.

1. Melissa’s POTW Solution

2. Because this is an equation both sides have to be equalSince the x3 gotten in a(x-1)3 =a (x3-3x2+3x-1) is the only term of that degree on the right side of the equation a=1 because the X3*1 = X3. 1(x3-3x2+3x-1) = (x3-3x2+3x-1) b(x-1)2= b(x2-2x+1) All the x2 term’s coefficients on the left side must add up to 5 that -3+b = 5 allows you to solve for b. b=8 Compute the value of c such that x3+5x2-8x+3= a(x-1)3+b(x-1)2+c(x-1)+d

3. x3-3x2+3x-1+b(x2-2x+1) After finding b plug it back x3-3x2+3x-1+8(x2-2x+1) into the equation. x3-3x2+3x-1 + 8x2-16x+8 x3+5x2 -13x+7 -13+c=-8 Using the x term values from the c=5 othertwo terms and -8 you can solve for c. Compute the value of c such that x3+5x2-8x+3= a(x-1)3+b(x-1)2+c(x-1)+d