How do you find the quotient of $\frac{t^{4} + 6 t^{3} - 3 t^{2} - t + 8}{t + 1}$ $(t^4+6t^3-3t^2-t+8)/(t+1)$ using long division?

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How do you find the quotient of $\frac{t^{4} + 6 t^{3} - 3 t^{2} - t + 8}{t + 1}$ $(t^4+6t^3-3t^2-t+8)/(t+1)$ using long division?

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Answer 1 · 2016-12-31T07:14:47

The quotient is $= t^{3} + 5 t^{2} - 8 t + 7$ $=t^3+5t^2-8t+7$ and the remainder is $= 1$ $=1$

Explanation:

Let's do the lomg division

$a a a a$ $color(white)(aaaa)$ $t^{4} + 6 t^{3} - 3 t^{2} - t + 8$ $t^4+6t^3-3t^2-t+8$ $a a a a$ $color(white)(aaaa)$ $∣$ $∣$ $t + 1$ $t+1$

$a a a a$ $color(white)(aaaa)$ $t^{4} + t^{3}$ $t^4+t^3$ $a a a a a a a a a a a a a a a a$ $color(white)(aaaaaaaaaaaaaaaa)$ $∣$ $∣$ $t^{3} + 5 t^{2} - 8 t + 7$ $t^3+5t^2-8t+7$

$a a a a$ $color(white)(aaaa)$ $0 + 5 t^{3} - 3 t^{2}$ $0+5t^3-3t^2$

$a a a a a a$ $color(white)(aaaaaa)$ $+ 5 t^{3} + 5 t^{2}$ $+5t^3+5t^2$

$a a a a a a a$ $color(white)(aaaaaaa)$ $+ 0 - 8 t^{2} - t$ $+0-8t^2-t$

$a a a a a a a a a a a$ $color(white)(aaaaaaaaaaa)$ $- 8 t^{2} - 8 t$ $-8t^2-8t$

$a a a a a a a a a a a a$ $color(white)(aaaaaaaaaaaa)$ $- 0 + 7 t + 8$ $-0+7t+8$

$a a a a a a a a a a a a a a a a$ $color(white)(aaaaaaaaaaaaaaaa)$ $+ 7 t + 7$ $+7t+7$

$a a a a a a a a a a a a a a a a a$ $color(white)(aaaaaaaaaaaaaaaaa)$ $+ 0 + 1$ $+0+1$

The quotient is $= t^{3} + 5 t^{2} - 8 t + 7$ $=t^3+5t^2-8t+7$ and the remainder is $= 1$ $=1$

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How do you find the quotient of $\frac{t^{4} + 6 t^{3} - 3 t^{2} - t + 8}{t + 1}$ $(t^4+6t^3-3t^2-t+8)/(t+1)$ using long division?

1 Answer

Explanation:

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How do you find the quotient of t4+6t3−3t2−t+8t+1(t^4+6t^3-3t^2-t+8)/(t+1) using long division?

1 Answer

Explanation:

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How do you find the quotient of $\frac{t^{4} + 6 t^{3} - 3 t^{2} - t + 8}{t + 1}$ $(t^4+6t^3-3t^2-t+8)/(t+1)$ using long division?