How do you divide $\frac{2 x^{3} - 11 x^{2} + 12 x + 9}{x - 3}$ $(2x^3 - 11x^2 + 12x + 9)/(x-3)$ ?

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How do you divide $\frac{2 x^{3} - 11 x^{2} + 12 x + 9}{x - 3}$ $(2x^3 - 11x^2 + 12x + 9)/(x-3)$ ?

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Answer 1 · 2018-07-27T05:56:33

The remainder is $0$ $0$ and the quotient is $= 2 x^{2} - 5 x - 3$ $=2x^2-5x-3$

Explanation:

Let's perform the synthetic division

$a a a a$ $color(white)(aaaa)$ $3$ $3$ $∣$ $|$ $a a a a$ $color(white)(aaaa)$ $2$ $2$ $a a a a$ $color(white)(aaaa)$ $- 11$ $-11$ $a a a a a a$ $color(white)(aaaaaa)$ $12$ $12$ $a a a a a$ $color(white)(aaaaa)$ $9$ $9$

$a a a a a$ $color(white)(aaaaa)$ $∣$ $|$ $a a a a$ $color(white)(aaaa)$ $a a a a a a a$ $color(white)(aaaaaaa)$ $6$ $6$ $a a a a a$ $color(white)(aaaaa)$ $- 15$ $-15$ $a a a$ $color(white)(aaa)$ $- 9$ $-9$

$a a a a a a a a a$ $color(white)(aaaaaaaaa)$ _________

$a a a a a$ $color(white)(aaaaa)$ $∣$ $|$ $a a a a$ $color(white)(aaaa)$ $2$ $2$ $a a a a a$ $color(white)(aaaaa)$ $- 5$ $-5$ $a a a a a$ $color(white)(aaaaa)$ $- 3$ $-3$ $a a a a a$ $color(white)(aaaaa)$ $0$ $color(red)(0)$

The remainder is $0$ $0$ and the quotient is $= 2 x^{2} - 5 x - 3$ $=2x^2-5x-3$

$\frac{2 x^{3} - 11 x^{2} + 12 x + 9}{x - 3} = 2 x^{2} - 5 x - 3$ $(2x^3-11x^2+12x+9)/(x-3)=2x^2-5x-3$

Apply the remainder theorem

When a polynomial $f (x)$ $f(x)$ is divided by $(x - c)$ $(x-c)$ , we get

$f (x) = (x - c) q (x) + r$ $f(x)=(x-c)q(x)+r$

Let $x = c$ $x=c$

Then,

$f (c) = 0 + r$ $f(c)=0+r$

Here,

$f (x) = 2 x^{3} - 11 x^{2} + 12 x + 9$ $f(x)=2x^3-11x^2+12x+9$

Therefore,

$f (3) = 2 \cdot 3^{3} - 11 \cdot 3^{2} + 12 \cdot 3 + 9$ $f(3)=2*3^3-11*3^2+12*3+9$

$= 54 - 99 + 36 + 9$ $=54-99+36+9$

$= 0$ $=0$

The remainder is $= 0$ $=0$

Answer 2 · 2018-07-27T05:58:00

$(2 x^{3} - 11 x^{2} + 12 x + 9) = (x - 3) (2 x^{2} - 5 x - 3) + (0)$ $(2x^3-11x^2+12x+9)=(x-3)(2x^2-5x-3)+(0)$

Explanation:

$(2 x^{3} - 11 x^{2} + 12 x + 9) \div (x - 3)$ $(2x^3-11x^2+12x+9)div(x-3)$

We can divide this polynomial by using synthetic division

We have , $p (x) = 2 x^{3} - 11 x^{2} + 12 x + 9 and divisor : x = 3$ $p(x)=2x^3-11x^2+12x+9 and "divisor :"x=3$

We take ,coefficients of $p (x) \to 2, - 11, 12, 9$ $p(x) to 2,-11,12 ,9$

. $3 ∣$ $3 |$ $2color(white)(....)-11color(white)(.......)12color(white)(.......)9$
$ulcolor(white)()|$ $ul(0color(white)( ..........)6color(white)(....)-15color(white)(...)-9$
$color(white)(......)2color(white)(......)-5color(white)(...)color(white)(..)-3color(white)(.....)color(violet)(ul|0|$
We can see that , quotient polynomial :

$q(x)=2x^2-5x-3 and"the Remainder"=0$

Hence ,

$(2x^3-11x^2+12x+9)=(x-3)(2x^2-5x-3)+(0)$

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How do you divide $\frac{2 x^{3} - 11 x^{2} + 12 x + 9}{x - 3}$ $(2x^3 - 11x^2 + 12x + 9)/(x-3)$ ?

2 Answers

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