Question 3.3.4: Finding a Polynomial Function That Satisfies Given Condition...
Finding a Polynomial Function That Satisfies Given Conditions (Real Zeros)
Find a polynomial function ƒ(x) of degree 3 with real coefficients that satisfies the given conditions.
(a) Zeros of -1, 2, and 4; ƒ(1) = 3
(b) -2 is a zero of multiplicity 3; ƒ(-1) = 4
(a) These three zeros give x – (-1) = x + 1, x – 2, and x – 4 as factors of ƒ(x). Because ƒ(x) is to be of degree 3, these are the only possible factors by the number of zeros theorem. Therefore, ƒ(x) has the form
ƒ(x) = a(x + 1)(x – 2)(x – 4), for some real number a.
To find a, use the fact that ƒ(1) = 3.
ƒ(1) = a(1 + 1)(1 – 2)(1 – 4) Let x = 1.
3 = a(2)(-1)(-3) ƒ(1) = 3
3 = 6a Multiply.
a =\frac{1}{2} Divide by 6.
Thus, ƒ(x) = \frac{1}{2} (x + 1)(x – 2)(x – 4) , \text{Let} a = \frac{1}{2}.
or, ƒ(x) = \frac{1}{2} x³ – \frac{5}{2} x² + x + 4. Multiply.
(b) The polynomial function ƒ(x) has the following form.
ƒ(x) = a(x + 2)(x + 2)(x + 2) Factor theorem
ƒ(x) = a(x + 2)³ (x + 2) is a factor three times.
To find a, use the fact that ƒ(-1) = 4.
ƒ(-1) = a(-1 + 2)³ Let x = -1.
4 = a(1)³ ƒ(-1) = 4
a = 4 Solve for a.
Thus, ƒ(x) = 4(x + 2)³,
or, ƒ(x)= 4x³ + 24x² + 48x + 32. Multiply.