How do you simplify and divide $(n^{3} + 2 n^{2} - 5 n + 12) \div (n + 4)$ $(n^3+2n^2-5n+12)div(n+4)$ ?

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How do you simplify and divide $(n^{3} + 2 n^{2} - 5 n + 12) \div (n + 4)$ $(n^3+2n^2-5n+12)div(n+4)$ ?

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Answer 1 · 2016-12-01T10:32:14

The remainder is $= 0$ $=0$
and the quotient is $= n^{2} - 2 n + 3$ $=n^2-2n+3$

Explanation:

Let's do the long division

$a a a a$ $color(white)(aaaa)$ $n^{3} + 2 n^{2} - 5 n + 12$ $n^3+2n^2-5n+12$ $a a a a$ $color(white)(aaaa)$ $∣$ $∣$ $n + 4$ $n+4$

$a a a a$ $color(white)(aaaa)$ $n^{3} + 4 n^{2}$ $n^3+4n^2$ $a a a a a a a a a a a a a$ $color(white)(aaaaaaaaaaaaa)$ $∣$ $∣$ $n^{2} - 2 n + 3$ $n^2-2n+3$

$a a a a a$ $color(white)(aaaaa)$ $0 - 2 n^{2} - 5 n$ $0-2n^2-5n$

$a a a a a a a$ $color(white)(aaaaaaa)$ $- 2 n^{2} - 8 n$ $-2n^2-8n$

$a a a a a a a a a a a$ $color(white)(aaaaaaaaaaa)$ $0 + 3 n + 12$ $0+3n+12$

$a a a a a a a a a a a a a$ $color(white)(aaaaaaaaaaaaa)$ $+ 3 n + 12$ $+3n+12$

$a a a a a a a a a a a a a a a$ $color(white)(aaaaaaaaaaaaaaa)$ $+ 0 + 0$ $+0+0$

So, the remainder is $= 0$ $=0$ and the quotient is $= n^{2} - 2 n + 3$ $=n^2-2n+3$

If we use the remainder theorem

$f (n) = n^{3} + 2 n^{2} - 5 n + 12$ $f(n)= n^3+2n^2-5n+12$

$f (- 4) = - 64 + 32 + 20 + 12 = 0$ $f(-4)=-64+32+20+12=0$

The remainder is $= 0$ $=0$

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How do you simplify and divide $(n^{3} + 2 n^{2} - 5 n + 12) \div (n + 4)$ $(n^3+2n^2-5n+12)div(n+4)$ ?

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Explanation:

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How do you simplify and divide (n^3+2n^2-5n+12)div(n+4)(n3+2n2−5n+12)÷(n+4)(n^3+2n^2-5n+12)div(n+4)?

1 Answer

Explanation:

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How do you simplify and divide $(n^{3} + 2 n^{2} - 5 n + 12) \div (n + 4)$ $(n^3+2n^2-5n+12)div(n+4)$ ?