# When the polynomial x^4 + bx^3 + 5x^2 + dx + 6 is divided by x - 2 the remainder is 16. When it is divided by x + 1 the remainder is 10. Find the value of constant d.

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**Problem**

When the polynomial $x^4 + bx^3 + 5x^2 + dx + 6$ is divided by $x - 2$ the remainder is 16. When it is divided by $x + 1$ the remainder is 10. Find the value of constant $d$.

A. 7 | C. -5 |

B. -7 | D. 5 |

**Answer Key**

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[ A ]

**Solution**

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$f(x) = x^4 + bx^3 + 5x^2 + dx + 6$

When

$\color{#FC6255}{f(2) =} 2^4 + b(2^3) + 5(2^2) + d(2) + 6 = 16$

By Remainder Theorem

*f*(

*x*) is divided by

*x*- 2, the remainder is

*f*(2)

$\color{#FC6255}{f(2) =} 2^4 + b(2^3) + 5(2^2) + d(2) + 6 = 16$

$8b + 2d = -26$ ← Equation (1)

When *f*(*x*) is divided by *x* + 1, the remainder is *f*(-1)

$\color{#FC6255}{f(-1) =} (-1)^4 + b(-1)^3 + 5(-1)^2 + d(-1) + 6 = 10$

$-b - d = -2$ ← Equation (2)

From Equation (1) and Equation (2)

$b = -5$

$d = 7$ ← *answer*

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