How do you divide $3 x^{3} + 5 x^{2} + 2 x - 12$ $3x^3+5x^2+2x-12$ by $x + 3$ $x+3$ ?

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How do you divide $3 x^{3} + 5 x^{2} + 2 x - 12$ $3x^3+5x^2+2x-12$ by $x + 3$ $x+3$ ?

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Answer 1 · 2018-04-07T13:23:44

$p (x) = 3 x^{2} - 4 x + 14 + \frac{(- 54)}{x + 3}$ $p(x)=3x^2-4x+14+((-54))/(x+3)$

Explanation:

Let $3 x^{3} + 5 x^{2} + 2 x - 12$ $3x^3+5x^2+2x-12$ be $p (x)$ $p(x)$
Using long division of polynomial's
$(+) 3 x^{2} - 4 x + 14$ $(+)3x^2-4x+14$
$\sqrt[x + 3]{3 x^{3} + 5 x^{2} + 2 x - 12}$ $root(x+3)(3x^3+5x^2+2x-12)$
$\frac{(-) 3 x^{3} + 9 x^{2}}{- 4 x^{2} + 2 x - 12}$ $((-)3x^3+9x^2)/(-4x^2+2x-12)$
$\frac{(-) - 4 x^{2} - 12 x}{14 x - 12}$ $((-)-4x^2-12x)/(14x-12)$
$\frac{(-) 14 x + 42}{- 54}$ $((-)14x+42)/-54$
$p (x) = Q (x) + \frac{R}{D}$ $p(x)=Q(x)+(R)/D$
Where, $Q =$ $Q=$ quotient
$R =$ $R=$ remainder
$D =$ $D=$ divisor
$\Rightarrow p (x) = 3 x^{2} - 4 x + 14 + \frac{(- 54)}{x + 3}$ $=>p(x)=3x^2-4x+14+((-54))/(x+3)$

Answer 2 · 2018-04-07T13:45:17

$3 x^{2} - 4 x + 14 - \frac{54}{x + 3}$ $3x^2-4x+14-54/(x+3)$

Explanation:

$one way is to use the divisor as a factor in the numerator$ $"one way is to use the divisor as a factor in the numerator"$

$consider the numerator$ $"consider the numerator"$

$3 x^{2} (x + 3) - 9 x^{2} + 5 x^{2} + 2 x - 12$ $color(red)(3x^2)(x+3)color(magenta)(-9x^2)+5x^2+2x-12$

$= 3 x^{2} (x + 3) - 4 x (x + 3) + 12 x + 2 x - 12$ $=color(red)(3x^2)(x+3)color(red)(-4x)(x+3)color(magenta)(+12x)+2x-12$

$= 3 x^{2} (x + 3) - 4 x (x + 3) + 14 (x + 3) - 42 - 12$ $=color(red)(3x^2)(x+3)color(red)(-4x)(x+3)color(red)(+14)(x+3)color(magenta)(-42)-12$

$= 3 x^{2} (x + 3) - 4 x (x + 3) + 14 (x + 3) - 54$ $=color(red)(3x^2)(x+3)color(red)(-4x)(x+3)color(red)(+14)(x+3)-54$

$quotient = 3 x^{2} - 4 x + 14, remainder = - 54$ $"quotient "=color(red)(3x^2-4x+14)," remainder "=-54$

$\Rightarrow 3 x^{3} + 5 x^{2} + 2 x - 12$ $rArr3x^3+5x^2+2x-12$

$= 3 x^{2} - 4 x + 14 - \frac{54}{x + 3}$ $=3x^2-4x+14-54/(x+3)$

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How do you divide $3 x^{3} + 5 x^{2} + 2 x - 12$ $3x^3+5x^2+2x-12$ by $x + 3$ $x+3$ ?

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