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

`p\left((2)/(3)\right)=-(1280)/(27)`

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

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=(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}ccccc(2)/(3) &6&19&-2&-44&-24\\\cline{1-1}\end{array}`

Bring the leading coefficient straight down:

`\begin{array}ccccc(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(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(2)/(3) & 6 & 19 & -2 & -44& -24\\\cline{1-1}& \downarrow &4 & & & \\\cline{2-6}& 6&23\end{array}`

Repeat:

`\begin{array}crrrr(2)/(3) & 6 & 19 & -2 & -44& -24\\\cline{1-1}& \vphantom{\frac12}\downarrow &4 &(46)/(3) & & \\\cline{2-6} \vphantom{\frac12}& 6&23&(40)/(3)\end{array}`

`\begin{array}crrrr(2)/(3) & 6 & 19 & -2 & -44& -24\\\cline{1-1}& \vphantom{\frac12}\downarrow &4 &(46)/(3) & (80)/(9)& \\\cline{2-6}& \vphantom{\frac12}6&23&(40)/(3)&-(316)/(9)\end{array}`

`\begin{array}c(2)/(3) & 6 & 19 & -2 & -44& -24\\\cline{1-1}&\vphantom{\frac12} \downarrow &4 &(46)/(3) & (80)/(9)&-(632)/(27) \\\cline{2-6}& \vphantom{\frac12}6&23&(40)/(3)&-(316)/(9)&-(1280)/(27)\end{array}`

The last number (remainder) is

`-(1280)/(27)`

Therefore, according to the remainder theorem:

`p\left((2)/(3)\right)=-(1280)/(27)`

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

`\implies p\left((2)/(3)\right)=6\left((2)/(3)\right)^4+19\left((2)/(3)\right)^3-2\left((2)/(3)\right)^2-44\left((2)/(3)\right)-24`

`\implies p\left((2)/(3)\right)=(32)/(27)+(152)/(27)-(8)/(9)-(88)/(3)-24`

`\implies p\left((2)/(3)\right)=-(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}crrr-1 &1&3&-5&-4\\\cline{1-1}\end{array}`

Bring the leading coefficient straight down:

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

Repeat:

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

`\begin{array}c -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`