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Difficulty: HardPolynomial Factors and Graphs

A polynomial function pp of degree 3 has xx-intercepts at (−2,0)(-2, 0) with multiplicity 2, and (3,0)(3, 0) with multiplicity 1. In the xyxy-plane, the graph of y=p(x)y = p(x) intersects the yy-axis at (0,24)(0, 24). What is the remainder when p(x)p(x) is divided by x−1x - 1?

  1. A
    88
  2. B
    1616
  3. 3636Answer
  4. D
    −36-36

Answer

The remainder when the polynomial function is divided by x−1x - 1 is 36.
The correct answer is 36. A polynomial with a root at x=−2x = -2 of multiplicity 2 and a root at x=3x = 3 of multiplicity 1 has the form p(x)=a(x+2)2(x−3)p(x) = a(x + 2)^2(x - 3). Since the yy-intercept is (0,24)(0, 24), we solve p(0)=a(2)2(−3)=24p(0) = a(2)^2(-3) = 24 to find a=−2a = -2. Thus, the polynomial is p(x)=−2(x+2)2(x−3)p(x) = -2(x + 2)^2(x - 3). According to the Remainder Theorem, dividing p(x)p(x) by x−1x - 1 leaves a remainder of p(1)p(1). Substituting x=1x = 1 yields p(1)=−2(3)2(−2)=36p(1) = -2(3)^2(-2) = 36.

Step-by-Step Solution

1
Write the general form of the cubic polynomial using its roots and multiplicities.
p(x)=a(x+2)2(x−3)p(x) = a(x + 2)^2(x - 3)
Since there is an xx-intercept at x=−2x = -2 with multiplicity 2, (x+2)2(x + 2)^2 is a factor. Since there is an xx-intercept at x=3x = 3 with multiplicity 1, (x−3)(x - 3) is a factor. Here, aa is a constant coefficient.
2
Determine the value of the constant coefficient aa using the yy-intercept.
a=−2a = -2, so p(x)=−2(x+2)2(x−3)p(x) = -2(x + 2)^2(x - 3)
The graph intersects the yy-axis at (0,24)(0, 24), meaning p(0)=24p(0) = 24. Substituting x=0x = 0 gives p(0)=a(0+2)2(0−3)=−12ap(0) = a(0 + 2)^2(0 - 3) = -12a. Setting −12a=24-12a = 24 yields a=−2a = -2.
3
Apply the Remainder Theorem to find the required remainder.
The remainder is equal to p(1)p(1).
By the Remainder Theorem, the remainder when a polynomial p(x)p(x) is divided by x−cx - c is p(c)p(c). Here, the divisor is x−1x - 1, so we evaluate the polynomial at x=1x = 1.
4
Calculate the value of p(1)p(1).
p(1)=36p(1) = 36
Substituting x=1x = 1 into p(x)=−2(x+2)2(x−3)p(x) = -2(x + 2)^2(x - 3) gives p(1)=−2(1+2)2(1−3)=−2(9)(−2)=36p(1) = -2(1 + 2)^2(1 - 3) = -2(9)(-2) = 36.

Key Concept

Identifying a polynomial from its roots and multiplicities, solving for its leading coefficient using a given point, and applying the Remainder Theorem.
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Topics

Polynomial Factors and Graphs

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