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

Feb 11, 2016

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

Explanation:

The remainder of the synthetic substitution will be equal to $P \left(1\right)$.

The terms in the top row of the division will be the coefficients of the terms. Don't forget the missing $x$ term: $\textcolor{\mathmr{and} a n \ge}{4} {x}^{3} \textcolor{\mathmr{and} a n \ge}{- 5} {x}^{2} + \textcolor{\mathmr{and} a n \ge}{0} x + \textcolor{\mathmr{and} a n \ge}{3}$

$\left.\begin{matrix}\underline{1} \text{|" & 4 & -5 & " 0" & " 3" & " " \\ " " & ul" " & ul(" "color(red)4) & ulcolor(blue)(-1) & ulcolor(green)(-1) & + \\ " " & color(red)4 & color(blue)(-1) & color(green)(-1) & "|"mathbf(" 2") & " }\end{matrix}\right.$

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 $P \left(1\right) = 2$.

This also shows us that

$\frac{4 {x}^{3} - 5 {x}^{2} + 3}{x - 1} = 4 {x}^{2} - x - 1 + \frac{2}{x - 1}$