How do you use synthetic substitution to find x=1 for #P(x)=4x^3-5x^2+3#?

1 Answer
Feb 11, 2016

Answer:

#P(1)=2#. For a demonstration of how to show this, refer to the explanation below.

Explanation:

The remainder of the synthetic substitution will be equal to #P(1)#.

The terms in the top row of the division will be the coefficients of the terms. Don't forget the missing #x# term: #color(orange)(4)x^3color(orange)(-5)x^2+color(orange)0x+color(orange)3#

#{:(ul1"|",4,-5," 0"," 3"," "),(" ",ul" ",ul(" "color(red)4),ulcolor(blue)(-1),ulcolor(green)(-1),+),(" ",color(red)4,color(blue)(-1),color(green)(-1),"|"mathbf(" 2")," "):}#

The following is an explanation of the synthetic division:

Bring the original #4# term down. Then multiply it by #1# to get #4# again, which you bring up into the next column. Add the #4# to the #-5# up top to get #-1#, and repeat the pattern of multiplying by #1#, bringing to the next column and adding.

Since the remainder is #2#, we see that #P(1)=2#.

This also shows us that

#(4x^3-5x^2+3)/(x-1)=4x^2-x-1+2/(x-1)#