Answer to Question #150189 in Algebra for modou lamin

Question

Find the quotient and the remainder of the function
F(x) =
3x4 + 8x3 + 7x + 2
x + 2
3
i) long division (5marks)
ii) Synthetic division (3marks)
b. i) Factorize completely F(x) = x3 + 9x2 + 23x + 15 (4marks)
ii) Show if the F(x) = 3x4 +8x3 +7x2

Answer 1

i) long division

$$\begin{matrix} & \underline{3x^3+2x^2-4x+15} \\ x+2&)3x^4+8x^3+7x+2\ \ \ \\ - \end{matrix}$$ $$\begin{matrix} & \underline{3x^4+6x^3 }&&\ \ \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2x^3+0x^2+7x+2 \\ \end{matrix}$$ $$-\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$$ $$\begin{matrix} & \underline{2x^3+4x^2 }&& \\ & \ \ \ -4x^2+7x+2 \end{matrix}$$ $$-\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$$ $$\begin{matrix} & \underline{-4x^2-8x}&&& \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 15x+2 \end{matrix}$$ $$-\ \ \ \ \$$ $$\begin{matrix} &\ \ \ \ \ \ \ \ \ \ \ \ \underline{15x+30} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -28 \end{matrix}$$

$$\dfrac{3x^4+8x^3+7x+2}{x+2}=3x^3+2x^2-4x+15+\dfrac{-28}{x+2}$$

$Quotient=3x^3+2x^2-4x+15$

$Remainder=-28$

ii) Synthetic division

$$\def\arraystretch{1.5} \begin{array}{c:c} & x^4 & x^3 & x^2 & x^1 & x^0 \\ \hline -2 & 3 & 8 & 0 & 7 & 2 \\ & & -6 & -4 & 8 & -30 \\ \hdashline & 3 & 2 & -4 &15 & -28 \end{array}$$

$$\dfrac{3x^4+8x^3+7x+2}{x+2}=3x^3+2x^2-4x+15+\dfrac{-28}{x+2}$$

$Quotient=3x^3+2x^2-4x+15$

$Remainder=-28$

b) i)

Synthetic division

$$\def\arraystretch{1.5} \begin{array}{c:c} & x^3 & x^2 & x^1 & x^0 \\ \hline -1 & 1 & 9 & 23 & 15 \\ & & -1 & -8 & 8 & \\ \hdashline & 1 & 8 & 15 &0 \end{array}$$

$$\dfrac{x^3+9x^2+23x+15}{x+1}=x^2+8x+15$$

$Quotient=x^2+8x+15$

$Remainder=0$

$x^2+8x+15=(x+5)(x+3)$

$$x^3+9x^2+23x+15=(x+5)(x+3)(x+1)$$

ii) Show if the F(x) = 3x4 +8x3 +7x2

Synthetic division

$$\def\arraystretch{1.5} \begin{array}{c:c} & x^4 & x^3 & x^2 & x^1 & x^0 \\ \hline -2 & 3 & 8 & 7 & 0 & 0 \\ & & -6 & -4 & -6 & 12 \\ \hdashline & 3 & 2 & 3 &-6 & 12 \end{array}$$

$$\dfrac{3x^4+8x^3+7x^2}{x+2}=3x^3+2x^2+3x-6+\dfrac{12}{x+2}$$

$Quotient=3x^3+2x^2+3x-6$

$Remainder=12$