Exercise 90 Page 373

To divide the given polynomials using synthetic division, all the terms of the dividend must be present. Since there are no missing terms, we do not need to rewrite the polynomial. Remember that the general form of the synthetic division divisor must be $x-a.$ Therefore, we need to rewrite ours a little bit.

$$\begin{array}{c}a+1\iff a-(\text{-}1)\end{array}$$

Now we are ready to divide!

$\begin{array}{rl}\begin{array}{r}\text{-}1\\ \end{array}& \begin{array}{rrcc}6& 1& \text{-}1& 4\\ & & & \end{array}\end{array}$

Bring down the $\text{first }$ $\text{coefficient}$

$\begin{array}{rl}\begin{array}{r}\text{-}1\\ \end{array}& \begin{array}{rrcc}6& 1& \text{-}1& 4\\ & & & \end{array}\\ & \begin{array}{cccc}6& & & \end{array}\end{array}$

Multiply the $\text{coefficient}$ by the $\text{divisor}$

$\begin{array}{rl}\begin{array}{r}\text{-}1\\ \end{array}& \begin{array}{rrcc}6& 1& \text{-}1& 4\\ & \text{-}6& & \end{array}\\ & \begin{array}{ccc}6& & \end{array}\end{array}$

Add down

$\begin{array}{rl}\begin{array}{r}\text{-}1\\ \end{array}& \begin{array}{rrcc}6& 1& \text{-}1& 4\\ & \text{-}6& & \end{array}\\ & \begin{array}{ccc}6& \text{-}5& \end{array}\end{array}$

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Repeat the process for all of the coefficients

Multiply the $\text{coefficient}$ by the $\text{divisor}$

$\begin{array}{rl}\begin{array}{r}\text{-}1\\ \end{array}& \begin{array}{rrcc}6& 1& \text{-}1& 4\\ & \text{-}6& 5& \end{array}\\ & \begin{array}{ccc}6& \text{-}5& \end{array}\end{array}$

Add down

$\begin{array}{rl}\begin{array}{r}\text{-}1\\ \end{array}& \begin{array}{rrcc}6& 1& \text{-}1& 4\\ & \text{-}6& 5& \end{array}\\ & \begin{array}{cccc}6& \text{-}5& 4& \end{array}\end{array}$

Multiply the $\text{coefficient}$ by the $\text{divisor}$

$\begin{array}{rl}\begin{array}{r}\text{-}1\\ \end{array}& \begin{array}{rrcc}6& 1& \text{-}1& 4\\ & \text{-}6& 5& \text{-}4\end{array}\\ & \begin{array}{cccc}6& \text{-}5& 4& \end{array}\end{array}$

Add down

$\begin{array}{rl}\begin{array}{r}\text{-}1\\ \end{array}& \begin{array}{rrcc}6& 1& \text{-}1& 4\\ & \text{-}6& 5& \text{-}4\end{array}\\ & \begin{array}{ccc}6& \text{-}5& 40\end{array}\end{array}$

The quotient is a polynomial of degree $2,$ with the above coefficients. The remainder is $0.$

$$\begin{array}{cc}\underline{\text{Quotient}}& \underline{\text{Remainder}}\\ 6a^2-5a+4& 0\end{array}$$

Checking Our Answer

Checking the Answer

We can check our answer by multiplying the quotient by the divisor, and then adding the remainder. If the answer equals the dividend, it means our answer is correct. Let's do it!

$\left(6a^2-5a+4\right)(a+1)+0\stackrel{?}{=}6a^3+a^2-a+4$

Distr

Distribute $(a+1)$

$6a^2(a+1)-5a(a+1)+4(a+1)+0\stackrel{?}{=}6a^3+a^2-a+4$

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Simplify left-hand side

Distr

Distribute $6a^2$

$6a^3+6a^2-5a(a+1)+4(a+1)+0\stackrel{?}{=}6a^3+a^2-a+4$

Distr

Distribute $\text{-}5a$

$6a^3+6a^2-5a^2-5a+4(a+1)+0\stackrel{?}{=}6a^3+a^2-a+4$

Distr

Distribute $4$

$6a^3+6a^2-5a^2-5a+4a+4+0\stackrel{?}{=}6a^3+a^2-a+4$

AddSubTerms

Add and subtract terms

$6a^3+a^2-a+4=6a^3+a^2-a+4✓$