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Difficulty: MediumPolynomial Factors and Graphs

A cubic polynomial function ff has xx-intercepts at (2,0)(2, 0), (5,0)(5, 0), and (c,0)(c, 0), where cc is a positive constant. In the xyxy-plane, the graph of y=f(x)y = f(x) has a yy-intercept at (0,60)(0, 60). If f(1)=16f(1) = 16, what is the value of cc?

Answer: 3

Answer

The correct answer is 3.
The factored form of a cubic polynomial with xx-intercepts at x=2x = 2, x=5x = 5, and x=cx = c is f(x)=a(x−2)(x−5)(x−c)f(x) = a(x-2)(x-5)(x-c). Evaluating the function at the yy-intercept x=0x = 0 gives f(0)=a(−2)(−5)(−c)=−10ac=60f(0) = a(-2)(-5)(-c) = -10ac = 60, which simplifies to ac=−6ac = -6. Using the point (1,16)(1, 16) yields f(1)=a(1−2)(1−5)(1−c)=4a(1−c)=16f(1) = a(1-2)(1-5)(1-c) = 4a(1-c) = 16. Expanding this equation gives 4a−4ac=164a - 4ac = 16. Substituting ac=−6ac = -6 into this equation yields 4a−4(−6)=164a - 4(-6) = 16, which simplifies to 4a+24=164a + 24 = 16, resulting in a=−2a = -2. Finally, substituting a=−2a = -2 into ac=−6ac = -6 yields −2c=−6-2c = -6, so c=3c = 3.

Step-by-Step Solution

1
Write the polynomial in factored form using its xx-intercepts.
f(x)=a(x−2)(x−5)(x−c)f(x) = a(x-2)(x-5)(x-c)
By the factor theorem, if a polynomial has xx-intercepts at x=r1,r2,r3x = r_1, r_2, r_3, then (x−r1)(x-r_1), (x−r2)(x-r_2), and (x−r3)(x-r_3) are factors of the polynomial.
2
Use the yy-intercept (0,60)(0, 60) to find a relation between aa and cc.
ac=−6ac = -6
Since the yy-intercept is at (0,60)(0, 60), we substitute x=0x = 0 into the polynomial and set the expression equal to 6060: a(−2)(−5)(−c)=−10ac=60a(-2)(-5)(-c) = -10ac = 60, which gives ac=−6ac = -6.
3
Use the given point f(1)=16f(1) = 16 to set up a second equation.
4a(1-c) = 16
Substitute x=1x = 1 and f(1)=16f(1) = 16 into the factored form: a(1−2)(1−5)(1−c)=a(−1)(−4)(1−c)=4a(1−c)=16a(1-2)(1-5)(1-c) = a(-1)(-4)(1-c) = 4a(1-c) = 16.
4
Solve for the leading coefficient aa by substituting ac=−6ac = -6.
a=−2a = -2
Expanding the equation from Step 3 yields 4a−4ac=164a - 4ac = 16. Substituting ac=−6ac = -6 gives 4a−4(−6)=164a - 4(-6) = 16, which simplifies to 4a+24=164a + 24 = 16, so 4a=−84a = -8 and a=−2a = -2.
5
Solve for the constant cc.
c=3c = 3
Using the relation ac=−6ac = -6 and substituting a=−2a = -2 gives −2c=−6-2c = -6, which yields c=3c = 3.

Key Concept

Using the factor theorem to set up a cubic polynomial equation and solving for unknown parameters using given points.
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Topics

Polynomial Factors and Graphs

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