College Algebra

Question

Find the polynomial function f with real coefficients that has the given degree, zeros, and solution point.

Degree: 4

Zeros: -3,1 i

Solution point: f(0)= -12

Answer 1

convert zeros to factors

change the sign & stick an x in front

x=3, (x-3)

x=1, (x-1)

x=i, (x-i)

imaginary zeros come in conjugate pairs

x=-i, (x+i)

multiply the factors together, with an unknown coefficient

f(x) = a(x-3)(x-1)(x-i)(x+i)

= a(x^2-4x+3)(x^2+1)=-12

replace x with 0

f(0)= a(3)(1) = -12

solve for a

a = -12/3 =- 4

f(x) = -4(x^2-4x+3)(x^2+1)

f(x)= -4(x^4-4x^3+4x^2-4x+3)

f(x) = -4x^4 +16x^3 -16x^2 +16x -12 is a 4th degree polynomial function with real coefficients & the given zeros

degree = the largest exponent= 4 = the leading term's exponent

check the answer, plug in the zeros

f(0)=-12

f(1)=-4+16-16+16-12=0

f(3)=-4(3^4)+16(3)^3)-16(3^2)+16(3)-12

= -4(81) + 16(27) -16(9) + 16(3) -12

= -324 +21(16) -12

= -324+ 126+210 -12

=-324 +336-12

= 0

f(i)=-4(1) + 16(-i) -16(-1) +16(i) -12

= -4 -16i +16+16i -12

=0

f(-i)=-4(1) +16(-i)+16 -16(-i)-12

= -4 -16i +16+16i -12

=-4+16-12

=0