SOLUTION: Suppose a polynomial P(x) when divided by x-9 has a quotient of 4x^3+5x^2+7x+13 and a remainder of 88. Find P(9).

Question 375366: Suppose a polynomial P(x) when divided by x-9 has a quotient of 4x^3+5x^2+7x+13 and a remainder of 88. Find P(9).
Answer by CharlesG2(834) (Show Source): You can put this solution on YOUR website!
Suppose a polynomial P(x) when divided by x-9 has a quotient of 4x^3+5x^2+7x+13 and a remainder of 88. Find P(9).

quotient: 4x^3 + 5x^2 + 7x + 13
quotient -> how many times did x - 9 go into P(x)
divisor: x - 9
remainder: 88
dividend: P(x)

(x - 9)(4x^3 + 5x^2 + 7x + 13)
4x^4 + 5x^3 + 7x^2 + 13x - 36x^3 - 45x^2 - 63x - 117
4x^4 - 31x^3 - 38x^2 - 50x - 117

-117 + 88 = -29

check:
...........4x^3 + 5x^2 + 7x + 13
x - 9 --> 4x^4 - 31x^3 - 38x^2 - 50x - 29
..........4x^4 - 36x^3
..................5x^3 - 38x^2
..................5x^3 - 45x^2
..........................7x^2 - 50x
..........................7x^2 - 63x
.................................13x - 29
.................................13x - 117
.......................................88
P(x) = 4x^4 - 31x^3 - 38x^2 - 50x - 29
P(9) = 4 * 9^4 - 31 * 9^3 - 38 * 9^2 - 50 * 9 - 29
P(9) = 4 * 6561 - 31 * 729 - 38 * 81 - 450 - 29
P(9) = 26244 - 22599 - 3078 - 479
P(9) = 88

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