Test 3 (Lessons 5–6): Dividing Polynomials Solutions

Simplify.

$(3 x^{5} y^{3} + 15 x^{4} y^{2} - 21 x^{3} y) {(3 x^{2} y)}^{- 1}$

$\frac{3 x^{5} y^{3} + 15 x^{4} y^{2} - 21 x^{3} y}{3 x^{2} y} \frac{3 x^{5} y^{3}}{3 x^{2} y} + \frac{15 x^{4} y^{2}}{3 x^{2} y} - \frac{21 x^{3} y}{3 x^{2} y} x^{3} y^{2} + 5 x^{2} y - 7 x$

$x^{3} y^{2} + 5 x^{2} y - 7 x$

$2 x + 3 \sqrt{10 x^{2} + 27 x + 18}$

$2 x + 3 5 x + 6 10 x^{2} + 27 x + 18 - (10 x^{2} + 15 x) 12 x + 18 - (12 x + 18) 0$

5x + 6

Divide $(x^{4} - 32)$ by $(x - 1) .$

$x - 1 x^{3} + x^{2} + x + 1 x^{4} + 0 x^{3} + 0 x^{2} + 0 x - 32 - (x^{4} - x^{3}) x^{3} + 0 x^{2} - (x^{3} - x^{2}) x^{2} + 0 x - (x^{2} - x) x - 32 - (x - 1) - 31 - \frac{31}{x - 1}$

$x^{3} + x^{2} + x + 1 - \frac{31}{x - 1}$

$\frac{5 a^{3} c - 12 a^{2} c^{2} + 8 a c - 16}{4 a c}$

$\frac{5 a^{3} c}{4 a c} - \frac{12 a^{2} c^{2}}{4 a c} + \frac{8 a c}{4 a c} - \frac{16}{4 a c}$

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

Simplify using long division.
$(4 x^{4} - 3 x^{3} + 2 x^{2} - x + 5) {(x + 1)}^{- 1}$

$x + 1 4 x^{3} - 7 x^{2} + 9 x - 10 4 x^{4} - 3 x^{3} + 2 x^{2} - x + 5 - (4 x^{4} + 4 x^{3}) - 7 x^{3} + 2 x^{2} - (- 7 x^{3} - 7 x^{2}) 9 x^{2} - x - (9 x^{2} + 9 x) - 10 x + 5 - (- 10 x - 10) 15 + \frac{15}{x + 1}$

$4 x^{3} - 7 x^{2} + 9 x - 10 + \frac{15}{x + 1}$

Simplify using synthetic division.
$(24 x^{4} + 15 x^{3} - 7 x^{2} + x - 5) \div (x + 1)$

$24 x^{3} - 9 x^{2} + 2 x - 1 - \frac{4}{x + 1}$

Use synthetic substitution to find the value of n when $P (- 2) = - 9 for P (x) = 2 x^{3} + 5 x^{2} + n x - 7 .$

$\begin{array}{rcl} - 2 (n - 2) - 7 & = & - 9 \\ - 2 (n - 2) & = & - 2 \\ - 2 n + 4 & = & - 2 \\ - 2 n & = & - 6 \end{array}$

n = 3

Use synthetic substitution to determine if p(−2) is a factor of $x^{4} + 8 x^{3} - 23 x - 6$ .
Explain.

Sample:
The remainder is –8, meaning –2 is not a factor of the polynomial because it does not go into the expression evenly.

Power P is the product of voltage E and current I, P = EI.
The power was measured to be $(a^{3} + 10 a^{2} + 28 a + 15)$ and the current is (a + 5). Determine the voltage.

$P = (a^{3} + 10 a^{2} + 28 a + 15), I = (a + 5) P = E I a^{3} + 10 a^{2} + 28 a + 15 = E (a + 5) E = \frac{a^{3} + 10 a^{2} + 28 a + 15}{a + 5} a + 5 = 0 a = - 5 E (- 5) using synthetic substitution .$

$a^{2} - 5 a + 3$

The voltage is

$(a^{2} - 5 a + 3)$

A polynomial is divided by a binomial and the remainder is zero. What does this tell you about the relationship between the polynomial and binomial expressions?

Sample:
When the remainder is zero, it means the binomial is a factor of the polynomial expression.