Practice 2 Solutions

Find the quotient using synthetic division.

$(3 y^{3} + 17 y^{2} + 22 y + 8) \div (y + 4)$

$3 y^{2} + 5 y + 2$

$(9 x^{2} - 7 x - 40) \div (x + 5)$

$9 x - 52 + \frac{220}{x + 5}$

$(x^{4} - 3 x^{2} + x - 5) {(x + 1)}^{- 1}$

$x^{3} - x^{2} - 2 x + 3 - \frac{8}{x + 1}$

$2 y - 1^{} 4 y^{2} - 8 y + 3$

$\frac{(4 y^{2} - 8 y + 3) \div 2}{(2 y - 1) \div 2} = \frac{2 y^{2} - 4 y + \frac{3}{2}}{y - \frac{1}{2}}$

Example Solution

$2 y - 3$

$\frac{2 y^{3} - 5 y - 2}{y + 1}$

$2 y^{2} - 2 y - 3 + \frac{1}{y + 1}$

$(2 x^{3} + 13 x^{2} - x - 110) {(x - \frac{5}{2})}^{- 1}$

$2 x^{2} + 18 x + 44 or 2 (x^{2} + 9 x + 22)$

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

$3 x^{3} + 9 x^{2} + 25 x + 80 + \frac{238}{x - 3}$

Note

This is the incorrect solution that would occur if you do not include $0 x^{3}$ as you create your synthetic division problem.

$(3 x^{2} + 4 x - 9) \div (3 x + 1)$

$\frac{(3 x^{2} + 4 x - 9) \div 3}{(3 x + 1) \div 3} = \frac{x^{2} + \frac{4}{3} x - 3}{x + \frac{1}{3}}$

$x + 1 - \frac{\frac{10}{3}}{x + \frac{1}{3}} \frac{(\frac{10}{3}) (3)}{(x + \frac{1}{3}) (3)} x + 1 - \frac{10}{3 x + 1}$ $x + 1 - \frac{10}{3 x + 1}$

Use the Remainder Theorem to determine P(k).

$P (x) = 4 x^{3} + 5 x^{2} - 3; P (- 3)$

$P (- 3) = - 66$

$P (x) = x^{5} + 5 x^{4} - 10 x^{3} + 10 x^{2} - 5 x - 1; P (1)$

P(1) = 0

Note

x = 1 is a root of the polynomial. x − 1 is a factor of the polynomial

$P (x) = - 2 x^{2} + 5 x + 3; P (- \frac{1}{2})$

$P (- \frac{1}{2}) = 0$

Note

$x = - \frac{1}{2}$ is a root of the polynomial. $(x + \frac{1}{2}) o r (2 x + 1)$ is a factor of the polynomial.

Find the remainder using $(x + 7)$ for $P (x) = 9 x^{4} + 62 x^{3} - x^{2} + 22 x - 100$

$P (- 7) = 40$

Find the missing value.

$P (2) = - 1; P (x) = x^{4} - 3 x^{2} + 4 x + n$

$\begin{array}{rcl} n + 12 & = & - 1 \\ n & = & - 13 \end{array}$

$P (- 4) = 6; P (x) = 5 x^{3} + 21 x^{2} - n x + 7$

$\begin{array}{rcl} - 4 (- n - 4) + 7 & = & 6 \\ 4 n + 16 + 7 & = & 6 \\ 4 n + 23 & = & 6 \\ 4 n & = & - 17 \\ n & = & - \frac{17}{4} \end{array}$

$\begin{array}{rcl} n & = & - \frac{17}{4} \end{array}$