Question

Given 1+2i as a zero of f(x) = x^4 - 2x^3 - 4x^2 + 18x - 45, find all other zeros and write the complete factorization. Please show work if you can.

Answer 1

(x−3)(x+3)(x^2-2x+5)

Explanation:

f(x) = x^4 - 2x^3 - 4x^2 + 18x - 45

To factor the expression, solve the equation where it equals to 0.

x^4 - 2x^3 - 4x^2 + 18x - 45 = 0

By Rational Root Theorem, all rational roots of a polynomial are in the form p/q where p divides the constant term −45 and q divides the leading coefficient 1. List all candidates p/q.

±45,±15,±9,±5,±3,±1

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

x=3

By Factor theorem, x−k is a factor of the polynomial for each root k. Divide x^4 - 2x^3 - 4x^2 + 18x - 45 by x−3 to get x^3 + x^2 - x + 15. To factor the result, solve the equation where it equals to 0.

x^3 + x^2 - x + 15 = 0

By Rational Root Theorem, all rational roots of a polynomial are in the form p/q where p divides the constant term 15 and q divides the leading coefficient 1. List all candidates p/q.

±15,±5,±3,±1

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

x=−3

By Factor theorem, x−k is a factor of the polynomial for each root k. Divide x^3 + x^2 - x + 15 by x+3 to get x^2 - 2x + 5. To factor the result, solve the equation where it equals to 0.

x^2−2x+5=0

All equations of the form ax^2 + bx + c=0 can be solved using the quadratic formula: −b±√b2-4ac/2a. Substitute 1 for a, −2 for b, and 5 for c in the quadratic formula.

x=-(-2)+√(-2)^2 - 4 x 1 x 5 / 2

Do the calculations.

x=2+√-16 / 2

Polynomial x^2−2x+5 is not factored since it does not have any rational roots.

x^2−2x+5

Rewrite the factored expression using the obtained roots.

(x−3)(x+3)(x^2-2x+5)

Answer 2

• (x + 3)(x - 3)(x² - 2x + 5)

Explanation:

One of the zeros given:

• 1 + 2i

Since it is a complex number, its conjugate is also a zero:

• 1 - 2i

The product of two factors are:

• (x - (1 + 2i))(x - (1 - 2i)) =
• x² - x(1 + 2i + 1 - 2i) + (1 - 4i²) =
• x² - 2x + 5

We need two more zeros. Let them be m and n, then we have:

• (x - m)(x - n) = x² - (m +n)x + mn

The product of all factors:

• (x² - 2x + 5)(x² - (m+n)x + mn) = x⁴ - 2x³ - 4x² + 18x - 45
• x⁴- (m+n+2)x³ + (mn+5)x² - (2mn+5m+5n)x + 5mn = x⁴- 2x³- 4x²+ 18x- 45

Compare left and right sides and solve for m and n:

• m + n + 2 = 2 ⇒ m + n = 0
• mn + 5 = -4 ⇒ mn = -9
• 2mn + 5(m + n) = -18 ⇒ mn = -9

From the above equations we get:

• m = -n
• -n² = -9
• n = 3
• m = -3

The other zero's are 3 and -3

All of the factors:

• (x + 3)(x - 3)(x² - 2x + 5)