Question

use the remainder theorem to find the remainder when P(x) = x^4 - 9x^3 - 5x^2 - 3x + 4 is divided by x+3

Answer 1

Let's divide and find the remainder for the following division:

x^4 - 9x^3 - 5x^2 - 3x + 4 by x+3​

Step 1: x^3 + (-12x^3 - 5x^2 - 3x + 4)/x + 3

Step 2: (-12x^3 - 5x^2 - 3x + 4)/x + 3 = 12x^2 + (31x^2 - 3x + 4)/(x + 3)

Step 3: x^3 + 12x^2 + (31x^2 - 3x + 4)/(x + 3) = x^3 + 12x^2 + 31x + (-96x + 4)/(x + 3)

Step 4: x^3 + 12x^2 + 31x + (-96x + 4)/(x + 3) = x^3 + 12x^2 + 31x - 96 + 292/(x + 3)

Step 5: x^3 + 12x^2 + 31x - 96 + 292/(x + 3) We can't continue because we have an integer in the numerator and a variable in the denominator, therefore:

x^4 - 9x^3 - 5x^2 - 3x + 4 / x + 3 = ​x^3 + 12x^2 + 31x - 96 + 292/(x + 3) , where the remainder is: 292/(x + 3)