Chapter 2.3, Problem 49E

To determine

To write : the function of the form $f\left(x\right)=\left(x-k\right)\cdot q\left(x\right)+r$ for the given value of k

Answer to Problem 49E

  $-4x^3+6x^2+12x+4=\boxed{\left(4x^2+\left(2+4\sqrt{3}\right)x+2\sqrt{3}+2\right)\left(x-1+\sqrt{3}\right)}$

Explanation of Solution

Given information : $f\left(x\right)=-4x^3+6x^2+12x+4,k=1-\sqrt{3}$

Concept Involved:

For long division of polynomials by divisors of the form x - k, there is a shortcut

called synthetic division. The pattern for synthetic division of a cubic polynomial is

summarized below. (The pattern for higher-degree polynomials is similar.)

Synthetic Division (for a Cubic Polynomial): To divide $a{x}^3+b{x}^2+cx+d$ by $x-k$ , use this pattern.

$\begin{array}{l}\begin{array}{l}k\\ \end{array}|\stackrel{Coefficientofdividend}{\overbrace{\underset{\_}{\begin{array}{cccc}\boxed{a}& b& c& d\\ \downarrow & \underset{\downarrow }{\boxed{ka}}& \underset{\downarrow }{\boxed{}}& \underset{\downarrow }{\boxed{}}\end{array}}}}\\ \begin{array}{ccccc}& \nearrow & \nearrow & \nearrow & \end{array}\\ \underset{Coefficientofquotient}{\underbrace{\begin{array}{ccc}\boxed{a}& \boxed{}& \boxed{}\end{array}}}\begin{array}{cc}\boxed{r}& \leftarrow \text{Remainder}\end{array}\end{array}$

In case when we have a polynomial with a missing term, insert placeholders with zero coefficients for missing powers of the variable. Vertical pattern: Add terms in columns Diagonal pattern: Multiply results by k. This algorithm for synthetic division works only for divisors of the form x - k. Remember that $x+k=x-\left(-k\right)$.

The Division Algorithm : If $f\left(x\right)$ and $d\left(x\right)$ are polynomials such that $d\left(x\right)\ne 0$ , and the degree of $d\left(x\right)$ isless than or equal to the degree of $f\left(x\right)$ , then there exist unique polynomials $q\left(x\right)$ and $r\left(x\right)$ such that $\underset{\stackrel{\uparrow }{{\displaystyle Dividend}}}{{\displaystyle \text{f (x)}}}\text{ = }\underset{\stackrel{\uparrow }{{\displaystyle Divisor}}}{{\displaystyle \text{d(x)}}}\cdot \underset{\stackrel{\uparrow }{{\displaystyle Quotient}}}{{\displaystyle \text{q(x)}}}+\underset{\stackrel{\uparrow }{{\displaystyle Remainder}}}{{\displaystyle \text{ r (x)}}}$ where $r\left(x\right)=0$ or the degree of $r\left(x\right)$ is less than the degree of $d\left(x\right)$ . If theremainder $r\left(x\right)$ is zero, then $d\left(x\right)$ divides evenly into $f\left(x\right)$ .

Calculation:

Set up the synthetic division to divide $f\left(x\right)=-4x^3+6x^2+12x+4,k=1-\sqrt{3}$ .

If $k=1-\sqrt{3}$ is zero, then $x-k$ that is $x-\left(1-\sqrt{3}\right)$ will be the divisor. When you set this up using synthetic division write $1-\sqrt{3}$ for the divisor. Then write the coefficients of the dividend to the right, across the top. Include any 0's that were inserted in for missing terms. In this polynomial there is no missing zeros. Write the terms of the dividend and divisor in descending powers of the variable.

  $\begin{array}{l}1-\sqrt{3}\\ \end{array}|\stackrel{Coefficientofdividend}{\overbrace{\underset{\_}{\begin{array}{cccc}-4& 6& 12& 4\\ & & & \end{array}}}}$

Bring down the leading coefficient to the bottom row.

  $\begin{array}{l}\begin{array}{l}1-\sqrt{3}\\ \end{array}|\underset{\_}{\begin{array}{cccc}-4& 6& 12& 4\\ \downarrow & & & \end{array}}\\ -4\end{array}$

Multiply $1-\sqrt{3}$ by the value just written on the bottom row. Place this value right beneath the next coefficient in the dividend:

  $\begin{array}{l}\begin{array}{l}1-\sqrt{3}\\ \searrow \end{array}|\underset{\_}{\begin{array}{cccc}-4& 6& 12& 4\\ & -4+4\sqrt{3}& & \end{array}}\\ -4^{\nearrow }\end{array}$

Add the column created and write the sum in the bottom row:

  $\begin{array}{l}\begin{array}{l}1-\sqrt{3}\\ \end{array}|\underset{\_}{\begin{array}{cccc}-4& 6& 12& 4\\ & -4+4\sqrt{3}\downarrow & & \end{array}}\\ -42+4\sqrt{3}\end{array}$

Repeat until done. Multiply $1-\sqrt{3}$ by the value just written on the bottom row. Place this value right beneath the next coefficient in the dividend:

  $\begin{array}{l}\begin{array}{l}1-\sqrt{3}\\ \searrow \end{array}|\underset{\_}{\begin{array}{cccc}-4& 6& 12& 4\\ & -4+4\sqrt{3}& 2\sqrt{3}-10& \end{array}}\\ -42+4{\sqrt{3}}^{\nearrow }\end{array}$

Add the column created and write the sum in the bottom row:

  $\begin{array}{l}\begin{array}{l}1-\sqrt{3}\\ \end{array}|\underset{\_}{\begin{array}{cccc}-4& 6& 12& 4\\ & -4+4\sqrt{3}& 2\sqrt{3}-10\downarrow & \end{array}}\\ -42+4\sqrt{3}2\sqrt{3}+2\end{array}$

Repeat until done. Multiply $1-\sqrt{3}$ by the value just written on the bottom row. Place this value right beneath the next coefficient in the dividend

  $\begin{array}{l}\begin{array}{l}1-\sqrt{3}\\ \searrow \end{array}|\underset{\_}{\begin{array}{cccc}-4& 6& 12& 4\\ & -4+4\sqrt{3}& 2\sqrt{3}-10& -4\end{array}}\\ -42+4\sqrt{3}2\sqrt{3}+2^{\nearrow }\end{array}$

Add the column created and write the sum in the bottom row:

  $\begin{array}{l}\begin{array}{l}1-\sqrt{3}\\ \end{array}|\underset{\_}{\begin{array}{cccc}-4& 6& 12& 4\\ & -4+4\sqrt{3}& 2\sqrt{3}-10& -4\downarrow \end{array}}\\ -42+4\sqrt{3}2\sqrt{3}+2\overline{)0}\end{array}$

Write out the answer.The numbers in the last row make up your coefficients of the quotient as well as the remainder. The final value on the right is the remainder. Working right to left, the next number is your constant, the next is the coefficient for x , the next is the coefficient for x squared, and so on.

  $\left(-4x^3+6x^2+12x+4\right)÷\left(x-1+\sqrt{3}\right)=4x^2+\left(2+4\sqrt{3}\right)x+\left(2\sqrt{3}+2\right)$

By dividing the given polynomial $-4x^3+6x^2+12x+4$ by $x-1+\sqrt{3}$ using the synthetic division we get the quotient as $4x^2+\left(2+4\sqrt{3}\right)x+\left(2\sqrt{3}+2\right)$ and the remainder as $0$ .

We can write the function in the form:

  $-4x^3+6x^2+12x+4=\left(4x^2+\left(2+4\sqrt{3}\right)x+2\sqrt{3}+2\right)\left(x-1+\sqrt{3}\right)$

Substituting $1-\sqrt{3}$ for $k$ in the function to show that $f\left(k\right)=r$ , in our problem $k=1-\sqrt{3}$ and we got remainder $r=0$

  $\begin{array}{l}f\left(1-\sqrt{3}\right)=-4{\left(1-\sqrt{3}\right)}^3+6{\left(1-\sqrt{3}\right)}^2+12\left(1-\sqrt{3}\right)+4\\ =-4\left(1-3\sqrt{3}+3\left(3\right)-3\sqrt{3}\right)+6\left(1-2\sqrt{3}+3\right)+12-12\sqrt{3}+4\\ =-4\left(10-6\sqrt{3}\right)+6\left(4-2\sqrt{3}\right)+16-12\sqrt{3}\\ =-40+24\sqrt{3}+24-12\sqrt{3}+16-12\sqrt{3}\\ =0\end{array}$