Question

Use synthetic division and remainder theorem to find P(a) P(x)
$p(x) = {6x}^{4} + {19x}^{3} - {2x}^{2} - 44x - 24$
a=2/3

$p(x) = {x}^{3} + {3x}^{2} - 5x - 4$
a=-1​

Answer 1

Answer:

$p\left(\dfrac{2}{3}\right)=-\dfrac{1280}{27}$

$p \left(-1 \right)=3$

Step-by-step explanation:

Remainder Theorem

When we divide a polynomial p(x) by (x − a) the remainder is p(a).

Given:

$\begin{cases}p(x)=6x^4+19x^3-2x^2-44x-24\\\\ a=\dfrac{2}{3}\end{cases}$

To find p(a), set up the synthetic division problem with the coefficients of the polynomial p(x) as the dividend and "a" as the divisor.

$\begin{array}{c|ccccc}\frac{2}{3} &6&19&-2&-44&-24\\\cline{1-1}\end{array}$

Bring the leading coefficient straight down:

$\begin{array}{c|ccccc}\frac{2}{3} & 6 & 19 & -2 & -44& -24\\\cline{1-1}& \downarrow & & & & \\\cline{2-6}& 6\end{array}$

Multiply the number you brought down with the number in the division box and put the result in the next column (under the 19):

$\begin{array}{c|ccccc}\frac{2}{3} & 6 & 19 & -2 & -44& -24\\\cline{1-1}& \downarrow &4 & & & \\\cline{2-6}& 6\end{array}$

Add the two numbers together and put the result in the bottom row:

$\begin{array}{c|crrrr}\frac{2}{3} & 6 & 19 & -2 & -44& -24\\\cline{1-1}& \downarrow &4 & & & \\\cline{2-6}& 6&23\end{array}$

Repeat:

$\begin{array}{c|crrrr}\frac{2}{3} & 6 & 19 & -2 & -44& -24\\\cline{1-1}& \vphantom{\dfrac12}\downarrow &4 &\frac{46}{3} & & \\\cline{2-6} \vphantom{\dfrac12}& 6&23&\frac{40}{3}\end{array}$

$\begin{array}{c|crrrr}\frac{2}{3} & 6 & 19 & -2 & -44& -24\\\cline{1-1}& \vphantom{\dfrac12}\downarrow &4 &\frac{46}{3} & \frac{80}{9}& \\\cline{2-6}& \vphantom{\dfrac12}6&23&\frac{40}{3}&-\frac{316}{9}\end{array}$

$\begin{array}{c|crrrr}\frac{2}{3} & 6 & 19 & -2 & -44& -24\\\cline{1-1}&\vphantom{\dfrac12} \downarrow &4 &\frac{46}{3} & \frac{80}{9}&-\frac{632}{27} \\\cline{2-6}& \vphantom{\dfrac12}6&23&\frac{40}{3}&-\frac{316}{9}&-\frac{1280}{27}\end{array}$

The last number (remainder) is

$-\dfrac{1280}{27}$

Therefore, according to the remainder theorem:

$p\left(\dfrac{2}{3}\right)=-\dfrac{1280}{27}$

Check by substituting a = 2/3 into p(x):

$\implies p\left(\dfrac{2}{3}\right)=6\left(\dfrac{2}{3}\right)^4+19\left(\dfrac{2}{3}\right)^3-2\left(\dfrac{2}{3}\right)^2-44\left(\dfrac{2}{3}\right)-24$

$\implies p\left(\dfrac{2}{3}\right)=\dfrac{32}{27}+\dfrac{152}{27}-\dfrac{8}{9}-\dfrac{88}{3}-24$

$\implies p\left(\dfrac{2}{3}\right)=-\dfrac{1280}{27}$

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Given:

$\begin{cases}p(x)=x^3+3x^2-5x-4\\ a=-1\end{cases}$

To find p(a), set up the synthetic division problem with the coefficients of the polynomial p(x) as the dividend and "a" as the divisor.

$\begin{array}{c|crrr}-1 &1&3&-5&-4\\\cline{1-1}\end{array}$

Bring the leading coefficient straight down:

$\begin{array}{c|crrr} -1 & 1&3&-5&-4\\\cline{1-1}& \downarrow & & & \\\cline{2-5}& 1\end{array}$

Multiply the number you brought down with the number in the division box and put the result in the next column (under the 3):

$\begin{array}{c|crrr} -1 & 1&3&-5&-4\\\cline{1-1}& \downarrow & -1& & \\\cline{2-5}& 1\end{array}$

Add the two numbers together and put the result in the bottom row:

$\begin{array}{c|crrr} -1 & 1&3&-5&-4\\\cline{1-1}& \downarrow & -1& & \\\cline{2-5}& 1&2\end{array}$

Repeat:

$\begin{array}{c|crrr} -1 & 1&3&-5&-4\\\cline{1-1}& \downarrow & -1& -2& \\\cline{2-5}& 1&2&-7\end{array}$

$\begin{array}{c|crrr} -1 & 1&3&-5&-4\\\cline{1-1}& \downarrow & -1& -2& 7\\\cline{2-5}& 1&2&-7&3\end{array}$

The last number (remainder) is 3.

Therefore, according to the remainder theorem:

$p \left(-1 \right)=3$

Check by substituting a = -1 into p(x):

$\implies p(-1)=(-1)^3+3(-1)^2-5(-1)-4$

$\implies p(-1)=-1+3+5-4$

$\implies p(-1)=3$