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For what value of b will the polynomial P(x) = 4x^3 - 3x^2 + bx + 6 have the same remainder when it is divided by x - 1 and by x +3?

1 Answer

1 vote

Answer:

29/7

Explanation:

If the polynomial P(x) has the same remainder when divided by x - 1 and by x + 3, then the difference between P(x) and its remainder when divided by x - 1 should be divisible by x - 1, and the difference between P(x) and its remainder when divided by x + 3 should be divisible by x + 3.

Let R1 and R2 be the remainders when P(x) is divided by x - 1 and by x + 3, respectively. Then, we have:

P(x) - R1 = (x - 1) Q(x) (1)

P(x) - R2 = (x + 3) S(x) (2)

where Q(x) and S(x) are polynomials.

Since P(x) is given as P(x) = 4x^3 - 3x^2 + bx + 6, we can substitute this expression into equations (1) and (2):

4x^3 - 3x^2 + bx + 6 - R1 = (x - 1) Q(x) (3)

4x^3 - 3x^2 + bx + 6 - R2 = (x + 3) S(x) (4)

To find the value of b that satisfies the problem condition, we need to set R1 = R2 and solve for b.

Subtracting equation (4) from equation (3), we get:

R1 - R2 = (x - 1) Q(x) - (x + 3) S(x)

Since R1 = P(1) and R2 = P(-3), we have:

P(1) - P(-3) = 4 - 3 + b(1 - 3) + 6 - (64 + 27 - 3b + 6)

Simplifying:

-57 + 4b = -28 - 3b

7b = 29

b = 29/7

Therefore, the value of b that will make P(x) have the same remainder when divided by x - 1 and by x + 3 is 29/7.

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