How do you divide $(6 b^{3} + 28 b^{2} + 25 b - 21) \div (b + 3)$ $(6b^3+28b^2+25b-21)div(b+3)$ using synthetic division?

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How do you divide $(6 b^{3} + 28 b^{2} + 25 b - 21) \div (b + 3)$ $(6b^3+28b^2+25b-21)div(b+3)$ using synthetic division?

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Answer 1 · 2017-02-21T13:56:09

The quotient is $= 6 b^{2} + 10 b - 5$ $=6b^2+10b-5$ and the remainder is $= - 6$ $=-6$

Explanation:

We perform the synthetic division

$a a a a$ $color(white)(aaaa)$ $- 3$ $-3$ $a a a a$ $color(white)(aaaa)$ $∣$ $|$ $a a a a$ $color(white)(aaaa)$ $6$ $6$ $a a a a$ $color(white)(aaaa)$ $28$ $28$ $a a a a a$ $color(white)(aaaaa)$ $25$ $25$ $a a a$ $color(white)(aaa)$ $- 21$ $-21$

$a a a a a a$ $color(white)(aaaaaa)$ $a a a a a$ $color(white)(aaaaa)$ $∣$ $|$ $a a a a a$ $color(white)(aaaaa)$ $a a$ $color(white)(aa)$ $- 18$ $-18$ $a a a a$ $color(white)(aaaa)$ $- 30$ $-30$ $a a a a$ $color(white)(aaaa)$ $15$ $15$

$a a a a a a a a a a$ $color(white)(aaaaaaaaaa)$ ------------------------------------------------------------

$a a a a$ $color(white)(aaaa)$ $a a a a a a$ $color(white)(aaaaaa)$ $a a a a a a$ $color(white)(aaaaaa)$ $6$ $6$ $a a a a$ $color(white)(aaaa)$ $10$ $10$ $a a a a$ $color(white)(aaaa)$ $- 5$ $-5$ $a a a a$ $color(white)(aaaa)$ $- 6$ $color(red)(-6)$

The remainder is $= - 6$ $=-6$

The quotient is $= 6 b^{2} + 10 b - 5$ $=6b^2+10b-5$

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How do you divide $(6 b^{3} + 28 b^{2} + 25 b - 21) \div (b + 3)$ $(6b^3+28b^2+25b-21)div(b+3)$ using synthetic division?

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Explanation:

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How do you divide (6b3+28b2+25b−21)÷(b+3)(6b^3+28b^2+25b-21)div(b+3) using synthetic division?

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Explanation:

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How do you divide $(6 b^{3} + 28 b^{2} + 25 b - 21) \div (b + 3)$ $(6b^3+28b^2+25b-21)div(b+3)$ using synthetic division?