How do you divide $(2 n^{3} + 62 n - 26 n^{2} + 4) \div (2 n - 6)$ $(2n^3+62n-26n^2+4)div(2n-6)$ using synthetic division?

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How do you divide $(2 n^{3} + 62 n - 26 n^{2} + 4) \div (2 n - 6)$ $(2n^3+62n-26n^2+4)div(2n-6)$ using synthetic division?

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Answer 1 · 2017-06-11T09:39:24

The remainder is $= 269$ $=269$ and the quotient is $= n^{2} + 34 n + 89$ $=n^2+34n+89$

Explanation:

Let's perform the synthetic division

$a a a a$ $color(white)(aaaa)$ $3$ $3$ $a a a a a$ $color(white)(aaaaa)$ $∣$ $|$ $a a a a$ $color(white)(aaaa)$ $1$ $1$ $a a a a a a$ $color(white)(aaaaaa)$ $31$ $31$ $a a a a a a$ $color(white)(aaaaaa)$ $- 13$ $-13$ $a a a a a$ $color(white)(aaaaa)$ $2$ $2$
$a a a a a a a a a a a a$ $color(white)(aaaaaaaaaaaa)$ _________

$a a a a$ $color(white)(aaaa)$ $a a a a a a a$ $color(white)(aaaaaaa)$ $∣$ $|$ $a a a a$ $color(white)(aaaa)$ $a a a a a a a$ $color(white)(aaaaaaa)$ $3$ $3$ $a a a a a a a$ $color(white)(aaaaaaa)$ $102$ $102$ $a a a a$ $color(white)(aaaa)$ $267$ $267$
$a a a a a a a a a a a a$ $color(white)(aaaaaaaaaaaa)$ ________

$a a a a$ $color(white)(aaaa)$ $a a a a a a a$ $color(white)(aaaaaaa)$ $∣$ $|$ $a a a a$ $color(white)(aaaa)$ $1$ $1$ $a a a a a a$ $color(white)(aaaaaa)$ $34$ $34$ $a a a a a a a$ $color(white)(aaaaaaa)$ $89$ $89$ $a a a a$ $color(white)(aaaa)$ $269$ $color(red)(269)$

The remainder is $= 269$ $=269$ and the quotient is $= n^{2} + 34 n + 89$ $=n^2+34n+89$

$\frac{n^{3} + 31 n^{2} - 13 n + 2}{n - 3} = n^{2} + 34 n + 89 + \frac{269}{n - 3}$ $(n^3+31n^2-13n+2)/(n-3)=n^2+34n+89+269/(n-3)$

N.B The numerator and denominator were divided by the common factor $2$ $2$

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How do you divide $(2 n^{3} + 62 n - 26 n^{2} + 4) \div (2 n - 6)$ $(2n^3+62n-26n^2+4)div(2n-6)$ using synthetic division?

1 Answer

Explanation:

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How do you divide (2n3+62n−26n2+4)÷(2n−6)(2n^3+62n-26n^2+4)div(2n-6) using synthetic division?

1 Answer

Explanation:

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How do you divide $(2 n^{3} + 62 n - 26 n^{2} + 4) \div (2 n - 6)$ $(2n^3+62n-26n^2+4)div(2n-6)$ using synthetic division?