Math

• Notes
• HW 2.6
• HW 2.7
• HW 3.1
• HW 3.2
• HW 3.3
• HW 3.4
• HW 3.5
• HW 4.1
• HW 4.2
• HW 4.3
• HW 4.4
• HW 4.5
• Exam November 9th
• Exam November 9th Corrections
• Test Review 03-08-2016

• Chem 7.31 - 7.45

HW 3.3 Q1

$(x^3 + 4x^2 + 8x + 5) \div (x+1)$

Long division:

$\space \space \space \space \space\space \space \space\space \space \space {x^2} + 3x + 5$
$x+1 \space \overline{)x^3 + 4x^2 + 8x + 5}$
$\space \space \space\space \space \space \space \space \space \space \space \space x^3 + x^2$
$\space \space \space \space \space \space \space \space \space\space \space \space\space \space \space\space \space \space \overline{3x^2 + 8x}$
$\space \space \space \space \space \space \space \space \space\space \space \space\space \space \space\space \space \space \overline{3x^2 + 3x}$
$\space \space \space \space \space \space \space \space \space\space \space \space\space \space \space\space \space \space \space \space \space \space \space \space \space \space \space \space \overline{5x + 5}$
$\space \space \space \space \space \space \space \space \space\space \space \space\space \space \space\space \space \space \space \space \space \space \space \space \space \space \space \space \overline{5x + 5}$

HW 3.3 Q2

$(8x^3 + 10x^2 - 9x - 20) \div (4x-5)$

Long division:

$\qquad \qquad 2x^2 +5x + 4$
$4x-5 \space \overline{) \space 8x^3 + 10x^2 - 9x - 20}$

$(8x^3 + 10x^2) - (8x^3 -10x^2) = 20x^2$

$( 20x^2 - 9x ) - ( 20x^2 -25x ) = 16x$

$(16x -20) - (16x-20) = 0$

HW 3.3 Q3

$(10x^2 - 3x - 16) \div (2x-3)$

Long division:

$\qquad \qquad 5x + 6$
$2x-3 \space \overline{) \space 10x^2 - 3x - 16}$

$(10x^2-3x) - (10x^2-15x) = 12x$

$(12x - 16) - (12x - 18) = 2$

$5x + 6 + \frac{2}{2x-3}$

HW 3.3 Q4

$(6x^3 - 9x^2 +7x + 4) \div (6x^2+3x+2)$

Long division:

$\qquad \qquad \qquad 1x - 2$
$6x^2+3x+2 \space \overline{) \space 6x^3 - 9x^2 +7x + 4}$

$(6x^3-9x^2+7x) - (6x^3+3x^2+2x) = -12x^2 +5x$

$(-12x^2+5x+4) - (-12x^2-6x-4) = 11x +8$

$x - 2 + \frac{11x +8}{6x^2+3x+2}$

HW 3.3 Q5

$(4x^4 +12x^3-2x^2) \div (2x^2+3)$

Long division:

$= 2x^2+6x-4 + \frac{-18x+12}{2x^2+3}$

HW 3.3 Q6

$(3x^2+3x+5) \div (x-3)$

Long division:

$= 3x+12 + \frac{41}{x-3}$

HW 3.3 Q9

$(5x^5-3x^3+3x^2-5x+7) \div (x-2)$

Synthetic division:

$= 5x^4 + 10x^3 + 17x^2 + 37x + 69 + \frac{145}{x-2}$

HW 3.3 Q16

(a)

$(14x^3 - 10x^2 + 12x - 96) = 0$

Show that $2$ is a solution of the equation.

Synthetic division:

$\color{blue}{14x^2 + 18x + 48 + \frac{0}{x-2}}$

(b)

$(14x^3 - 10x^2 + 12x - 3) = 93$

$14x^3 - 10x^2 + 12x - 3 - 93 = 0$
$= 14x^3 - 10x^2 + 12x - 96$

$x = 2$

$b^2 - 4ac$

$14x^2 + 18x + 48$
$a=14, b=18, c=48$

$b^2 - 4ac = 18^2 - (4)(14)(48) = 324 - 2688 = -2364$

Since the discriminate ($-2364$) is negative, the solution must be $x=2$

HW 3.3 Q17 (b)

$3h^3 + 21h^2 - 270 = 0$

$f(3) = 3x^2 + 30x + 90$

$v = l \times w \times h$
$v = 270$

$270 = (h+7) \times (3h) \times (h)$
$= (h+7) \times 3h^2 = 3h^3 + 21h^2$

$3h^3 + 21h^2 - 270 = 0$

$h = 3$
$w = 3h = 3(3) = 9$
$l = h+7 = 3+7 = 10$

$3 \times 9 \times 10 = 270$

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