How do you divide $(m^{3} + 6 m^{2} + 9 m - 5) \div (m + 1)$ $(m^3+6m^2+9m-5)div(m+1)$ using synthetic division?

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How do you divide $(m^{3} + 6 m^{2} + 9 m - 5) \div (m + 1)$ $(m^3+6m^2+9m-5)div(m+1)$ using synthetic division?

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Answer 1 · 2017-03-17T11:16:06

The remainder is $= - 9$ $=-9$ and the quotient is $= m^{2} + 5 m + 4$ $=m^2+5m+4$

Explanation:

Let's perform the synthetic division

$a a a a$ $color(white)(aaaa)$ $- 1$ $-1$ $a a a a$ $color(white)(aaaa)$ $∣$ $|$ $a a a a$ $color(white)(aaaa)$ $1$ $1$ $a a a a$ $color(white)(aaaa)$ $6$ $6$ $a a a a$ $color(white)(aaaa)$ $9$ $9$ $a a a a$ $color(white)(aaaa)$ $- 5$ $-5$
$a a a a a a a a a a a a$ $color(white)(aaaaaaaaaaaa)$ ____________

$a a a a$ $color(white)(aaaa)$ #color(white)(aaaaaaa)|# $a a a a$ $color(white)(aaaa)$ #color(white)(aaa)-1# $a a$ $color(white)(aa)$ $- 5$ $-5$ $a a a a$ $color(white)(aaaa)$ $- 4$ $-4$
$a a a a a a a a a a a a$ $color(white)(aaaaaaaaaaaa)$ ____________
$a a a a$ $color(white)(aaaa)$ #color(white)(aaaaaaa)|# $a a a a$ $color(white)(aaaa)$ $1$ $1$ $a a a a$ $color(white)(aaaa)$ $5$ $5$ $a a a a$ $color(white)(aaaa)$ $4$ $4$ $a a a a$ $color(white)(aaaa)$ $- 9$ $color(red)(-9)$

The remainder is $= - 9$ $=-9$ and the quotient is $= m^{2} + 5 m + 4$ $=m^2+5m+4$

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How do you divide $(m^{3} + 6 m^{2} + 9 m - 5) \div (m + 1)$ $(m^3+6m^2+9m-5)div(m+1)$ using synthetic division?

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Explanation:

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How do you divide (m3+6m2+9m−5)÷(m+1)(m^3+6m^2+9m-5)div(m+1) using synthetic division?

1 Answer

Explanation:

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How do you divide $(m^{3} + 6 m^{2} + 9 m - 5) \div (m + 1)$ $(m^3+6m^2+9m-5)div(m+1)$ using synthetic division?