Studdy AI Logo
Math  /  Algebra

QuestionFind the real zeros of f . Use the real zeros to factor f . f(x)=x4+10x3−20x2−90x+99f(x)=x^{4}+10 x^{3}-20 x^{2}-90 x+99

Studdy Solution

STEP 1

1. We are given the polynomial function f(x)=x4+10x3−20x2−90x+99 f(x) = x^4 + 10x^3 - 20x^2 - 90x + 99 .
2. We need to find the real zeros of f(x) f(x) .
3. Use the real zeros to factor f(x) f(x) .

STEP 2

1. Use the Rational Root Theorem to identify possible rational zeros.
2. Test possible rational zeros using synthetic division or direct substitution.
3. Confirm real zeros and factor the polynomial using the identified zeros.
4. Write the polynomial as a product of its factors.

STEP 3

Apply the Rational Root Theorem. The possible rational zeros are the factors of the constant term (99) divided by the factors of the leading coefficient (1).
Factors of 99: ±1,±3,±9,±11,±33,±99 \pm 1, \pm 3, \pm 9, \pm 11, \pm 33, \pm 99
Possible rational zeros: ±1,±3,±9,±11,±33,±99 \pm 1, \pm 3, \pm 9, \pm 11, \pm 33, \pm 99

STEP 4

Test possible rational zeros using synthetic division or substitution. Start with x=1 x = 1 :
f(1)=14+10(1)3−20(1)2−90(1)+99=1+10−20−90+99=0 f(1) = 1^4 + 10(1)^3 - 20(1)^2 - 90(1) + 99 = 1 + 10 - 20 - 90 + 99 = 0
Thus, x=1 x = 1 is a zero.

STEP 5

Perform synthetic division of f(x) f(x) by x−1 x - 1 :
1110−20−9099111−9−99111−9−990\begin{array}{r|rrrrr} 1 & 1 & 10 & -20 & -90 & 99 \\ & & 1 & 11 & -9 & -99 \\ \hline & 1 & 11 & -9 & -99 & 0 \\ \end{array}
The quotient is x3+11x2−9x−99 x^3 + 11x^2 - 9x - 99 .

STEP 6

Repeat the process to find more zeros of the quotient x3+11x2−9x−99 x^3 + 11x^2 - 9x - 99 .
Test x=−1 x = -1 :
f(−1)=(−1)3+11(−1)2−9(−1)−99=−1+11+9−99≠0 f(-1) = (-1)^3 + 11(-1)^2 - 9(-1) - 99 = -1 + 11 + 9 - 99 \neq 0
Test x=3 x = 3 :
f(3)=33+11(3)2−9(3)−99=27+99−27−99=0 f(3) = 3^3 + 11(3)^2 - 9(3) - 99 = 27 + 99 - 27 - 99 = 0
Thus, x=3 x = 3 is a zero.

STEP 7

Perform synthetic division of x3+11x2−9x−99 x^3 + 11x^2 - 9x - 99 by x−3 x - 3 :
3111−9−9934299114330\begin{array}{r|rrrr} 3 & 1 & 11 & -9 & -99 \\ & & 3 & 42 & 99 \\ \hline & 1 & 14 & 33 & 0 \\ \end{array}
The quotient is x2+14x+33 x^2 + 14x + 33 .

STEP 8

Factor the quadratic x2+14x+33 x^2 + 14x + 33 using the quadratic formula:
x=−b±b2−4ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} x=−14±142−4⋅1⋅332⋅1 x = \frac{-14 \pm \sqrt{14^2 - 4 \cdot 1 \cdot 33}}{2 \cdot 1} x=−14±196−1322 x = \frac{-14 \pm \sqrt{196 - 132}}{2} x=−14±642 x = \frac{-14 \pm \sqrt{64}}{2} x=−14±82 x = \frac{-14 \pm 8}{2}
The solutions are x=−3 x = -3 and x=−11 x = -11 .

STEP 9

Write the polynomial as a product of its factors using the zeros found:
f(x)=(x−1)(x−3)(x+3)(x+11) f(x) = (x - 1)(x - 3)(x + 3)(x + 11)
The real zeros of f(x) f(x) are x=1,3,−3,−11 x = 1, 3, -3, -11 , and the factored form of f(x) f(x) is:
f(x)=(x−1)(x−3)(x+3)(x+11) f(x) = (x - 1)(x - 3)(x + 3)(x + 11)

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord