Studdy AI Logo
Math  /  Algebra

QuestionFind the rational zeros and then the other zeros and factor into linear factors for the function f(x)=3x4−4x3+x2+6x−2 f(x) = 3x^{4} - 4x^{3} + x^{2} + 6x - 2 .

Studdy Solution

STEP 1

1. We are given the polynomial function f(x)=3x4−4x3+x2+6x−2 f(x) = 3x^4 - 4x^3 + x^2 + 6x - 2 .
2. We need to find all rational zeros of the polynomial.
3. We will then find the other zeros, if any.
4. Finally, we will factor the polynomial into linear factors.

STEP 2

1. Use the Rational Root Theorem to list all possible rational zeros.
2. Test each possible rational zero using synthetic division or direct substitution.
3. Once a rational zero is found, factor the polynomial using it.
4. Repeat the process to find all zeros.
5. Factor the polynomial completely into linear factors.

STEP 3

According to the Rational Root Theorem, the possible rational zeros of a polynomial are the factors of the constant term divided by the factors of the leading coefficient.
For f(x)=3x4−4x3+x2+6x−2 f(x) = 3x^4 - 4x^3 + x^2 + 6x - 2 , the constant term is −2-2 and the leading coefficient is 33.
Possible rational zeros are:
±1,±2,±13,±23 \pm 1, \pm 2, \pm \frac{1}{3}, \pm \frac{2}{3}

STEP 4

Test each possible rational zero using synthetic division or direct substitution to determine if it is a zero of the polynomial.
Let's start with x=1 x = 1 :
f(1)=3(1)4−4(1)3+(1)2+6(1)−2 f(1) = 3(1)^4 - 4(1)^3 + (1)^2 + 6(1) - 2 f(1)=3−4+1+6−2=4 f(1) = 3 - 4 + 1 + 6 - 2 = 4
Since f(1)≠0 f(1) \neq 0 , x=1 x = 1 is not a zero.

STEP 5

Continue testing other possible rational zeros.
Let's try x=−1 x = -1 :
f(−1)=3(−1)4−4(−1)3+(−1)2+6(−1)−2 f(-1) = 3(-1)^4 - 4(-1)^3 + (-1)^2 + 6(-1) - 2 f(−1)=3+4+1−6−2=0 f(-1) = 3 + 4 + 1 - 6 - 2 = 0
Since f(−1)=0 f(-1) = 0 , x=−1 x = -1 is a zero.

STEP 6

Use synthetic division to factor the polynomial by (x+1) (x + 1) .
Perform synthetic division of f(x) f(x) by x+1 x + 1 :
−13−416−2−37−823−78−20\begin{array}{r|rrrrr} -1 & 3 & -4 & 1 & 6 & -2 \\ & & -3 & 7 & -8 & 2 \\ \hline & 3 & -7 & 8 & -2 & 0 \\ \end{array}
The quotient is 3x3−7x2+8x−2 3x^3 - 7x^2 + 8x - 2 .

STEP 7

Now, find the zeros of the quotient polynomial 3x3−7x2+8x−2 3x^3 - 7x^2 + 8x - 2 .
Repeat the process of finding rational zeros for the new polynomial.
Possible rational zeros are:
±1,±2,±13,±23 \pm 1, \pm 2, \pm \frac{1}{3}, \pm \frac{2}{3}
Test x=1 x = 1 :
3(1)3−7(1)2+8(1)−2=3−7+8−2=2 3(1)^3 - 7(1)^2 + 8(1) - 2 = 3 - 7 + 8 - 2 = 2
Since f(1)≠0 f(1) \neq 0 , x=1 x = 1 is not a zero.

STEP 8

Test x=2 x = 2 :
3(2)3−7(2)2+8(2)−2=24−28+16−2=10 3(2)^3 - 7(2)^2 + 8(2) - 2 = 24 - 28 + 16 - 2 = 10
Since f(2)≠0 f(2) \neq 0 , x=2 x = 2 is not a zero.

STEP 9

Test x=−2 x = -2 :
3(−2)3−7(−2)2+8(−2)−2=−24−28−16−2=−70 3(-2)^3 - 7(-2)^2 + 8(-2) - 2 = -24 - 28 - 16 - 2 = -70
Since f(−2)≠0 f(-2) \neq 0 , x=−2 x = -2 is not a zero.

STEP 10

Test x=13 x = \frac{1}{3} :
3(13)3−7(13)2+8(13)−2=19−79+83−2 3\left(\frac{1}{3}\right)^3 - 7\left(\frac{1}{3}\right)^2 + 8\left(\frac{1}{3}\right) - 2 = \frac{1}{9} - \frac{7}{9} + \frac{8}{3} - 2
Simplifying:
=1−7+24−189=09=0 = \frac{1 - 7 + 24 - 18}{9} = \frac{0}{9} = 0
Since f(13)=0 f\left(\frac{1}{3}\right) = 0 , x=13 x = \frac{1}{3} is a zero.

STEP 11

Use synthetic division to factor the polynomial by (3x−1) (3x - 1) .
Perform synthetic division of 3x3−7x2+8x−2 3x^3 - 7x^2 + 8x - 2 by 3x−1 3x - 1 :
133−78−21−223−660\begin{array}{r|rrrr} \frac{1}{3} & 3 & -7 & 8 & -2 \\ & & 1 & -2 & 2 \\ \hline & 3 & -6 & 6 & 0 \\ \end{array}
The quotient is 3x2−6x+6 3x^2 - 6x + 6 .

STEP 12

Factor the quadratic 3x2−6x+6 3x^2 - 6x + 6 .
The quadratic can be factored as:
3(x2−2x+2) 3(x^2 - 2x + 2)
This quadratic does not have real roots, so we use the quadratic formula:
x=−b±b2−4ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
For x2−2x+2 x^2 - 2x + 2 , a=1,b=−2,c=2 a = 1, b = -2, c = 2 :
x=−(−2)±(−2)2−4⋅1⋅22⋅1 x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} x=2±4−82 x = \frac{2 \pm \sqrt{4 - 8}}{2} x=2±−42 x = \frac{2 \pm \sqrt{-4}}{2} x=2±2i2 x = \frac{2 \pm 2i}{2} x=1±i x = 1 \pm i
The complete factorization of the polynomial is:
f(x)=(x+1)(3x−1)(x−(1+i))(x−(1−i)) f(x) = (x + 1)(3x - 1)(x - (1 + i))(x - (1 - i))

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