Divide x³+3x²-x-2 by (x+3)(x+5)
The handout only teaches me how to do the remainder theorem with a linear divisor but doesn't mention anything about quadratic divisors. How do you use the remainder theorem with a quadratic divisor?
The handout only teaches me how to do the remainder theorem with a linear divisor but doesn't mention anything about quadratic divisors. How do you use the remainder theorem with a quadratic divisor?
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Let p(x) = x³ + 3x² - x - 2.
p(-3) = 1
p(-5) = -47
Let p(x) = (x + 3)(x + 5)q(x) + Ax + B, where q(x) is a polynomial.
p(-3) = -3A + B = 1
p(-5) = -5A + B = -47
A = 24
B = 73
When p(x) is divided by (x + 3)(x + 5), the remainder is 24x + 73.
p(-3) = 1
p(-5) = -47
Let p(x) = (x + 3)(x + 5)q(x) + Ax + B, where q(x) is a polynomial.
p(-3) = -3A + B = 1
p(-5) = -5A + B = -47
A = 24
B = 73
When p(x) is divided by (x + 3)(x + 5), the remainder is 24x + 73.
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Use FOIL to make (x+3)(x+5) a quadratic expression. (x+3)(x+5) = x^2 + 8x + 15. Next, find out how many times you need to multiply x^2 by to get the first term, x^3. Since x^2 goes into x^3 "x" times, multiply the quadratic expression by "x" to get x^3 + 8x^2 + 15x. Subtract this from the original dividend to get -5x^2 - 16x - 2. To get from x^2 to -5x^2, you need to multiply by -5 (the next part of you quotient). Multiplying x^2 + 8x + 15 by -5 results in -5x^2 - 40x - 75. Subtracting this from -5x^2 - 16x - 2 results in 24x + 73. As x^2 can't go into an x term, 24x + 73 is your remainder. You final answer is x-5 with a remainder of 24x+73.