Big Ideas Math Algebra 2, 2014
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Big Ideas Math Algebra 2, 2014 View details
5. Solving Polynomial Equations
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Exercise 6 Page 192

Start by looking for integer zeros. Integer zeros are factors of the constant term.

1, - 1/3, 2sqrt(3), and - 2sqrt(3)

Practice makes perfect

We want to find all real zeros of the given polynomial function. To do so, we need to solve the equation f(x)=0. 3x^4-2x^3-37x^2+24x+12=0 The degree of f(x) is 4. Thus, by the Fundamental Theorem of Algebra, we know that f(x)=0 has exactly four roots. Let's find them.

Integer Roots

By the Rational Root Theorem, we know that integer roots must be factors of the constant term. Since the constant term of f(x) is 12, the possible integer roots are ± 1, ± 2, ± 3, ± 4, ± 6, and ± 12. Let's check.

x 3x^4-2x^3-37x^2+24x+12 f(x)=3x^4-2x^3-37x^2+24x+12
1 3( 1)^4-2( 1)^3-37( 1)^2+24( 1)+12 0 âś“
- 1 3( - 1)^4-2( - 1)^3-37( - 1)^2+24( - 1)+12 - 44 *
2 3( 2)^4-2( 2)^3-37( 2)^2+24( 2)+12 - 56 *
- 2 3( - 2)^4-2( - 2)^3-37( - 2)^2+24( - 2)+12 - 120 *
3 3( 3)^4-2( 3)^3-37( 3)^2+24( 3)+12 - 60 *
- 3 3( - 3)^4-2( - 3)^3-37( - 3)^2+24( - 3)+12 - 96 *
4 3( 4)^4-2( 4)^3-37( 4)^2+24( 4)+12 156 *
- 4 3( - 4)^4-2( - 4)^3-37( - 4)^2+24( - 4)+12 220 *
6 3( 6)^4-2( 6)^3-37( 6)^2+24( 6)+12 2280 *
- 6 3( - 6)^4-2( - 6)^3-37( - 6)^2+24( - 6)+12 2856 *
12 3( 12)^4-2( 12)^3-37( 12)^2+24( 12)+12 53 724 *
- 12 3( - 12)^4-2( - 12)^3-37( - 12)^2+24( - 12)+12 60 060 *
We found that 1 is root for f(x)=0. Therefore, (x-1) is factor of the polynomial. Let's use synthetic division to factor out this factor and thus find the rest of the roots.
rl IR-0.15cm r 1 & |rr 3 &-2 &-37 & 24 & 12

Bring down the first coefficient

rl IR-0.15cm r 1 & |rr 3 &-2 &-37 & 24 & 12 &&&& & c 3 & & & &

Multiply the coefficient by the divisor

rl IR-0.15cm r 1 & |rr 3 &-2 &-37 & 24 & 12 & 3 &&& & c 3 & & & &

Add down

rl IR-0.15cm r 1 & |rr 3 &-2 &-37 & 24 & 12 & 3 &&& & c 3 & 1 & & &
â–Ľ
Repeat the process for all the coefficients

Multiply the coefficient by the divisor

rl IR-0.15cm r 1 & |rr 3 &-2 &-37 & 24 & 12 & 3 & 1 && & c 3 & 1 & & &

Add down

rl IR-0.15cm r 1 & |rr 3 &-2 &-37 & 24 & 12 & 3 & 1 && & c 3 & 1 &-36 & &

Multiply the coefficient by the divisor

rl IR-0.15cm r 1 & |rr 3 &-2 &-37 & 24 & 12 & 3 & 1 &-36 & & c 3 & 1 & -36 & &

Add down

rl IR-0.15cm r 1 & |rr 3 &-2 &-37 & 24 & 12 & 3 & 1 &-36 & & c 3 & 1 &-36 &-12 &

Multiply the coefficient by the divisor

rl IR-0.15cm r 1 & |rr 3 &-2 &-37 & 24 & 12 & 3 & 1 &-36 &-12 & c 3 & 1 &-36 & -12 &

Add down

rl IR-0.15cm r 1 & |rr 3 &-2 &-37 & 24 & 12 & 3 & 1 &-36 &-12 & c 3 & 1 &-36 &-12 & 0
Removing the first root, 1, gave us a smaller polynomial. 3x^3+x^2-36x-12

Rational Roots

By the Rational Root Theorem, we know that rational roots have the form ± pq, where p is an integer factor of the constant term, and q is an integer factor of the leading coefficient.

x 3x^4-2x^3-37x^2+24x+12 f(x)=3x^4-2x^3-37x^2+24x+12
1/3 3 ( 1/3 ) ^4-2 ( 1/3 ) ^3-37 ( 1/3 ) ^2+24 ( 1/3 ) +12 428/27 *
- 1/3 3 ( - 1/3 ) ^4-2 ( - 1/3 ) ^3-37 ( - 1/3 ) ^2+24 ( - 1/3 ) +12 0 âś“
We found that - 13 is a zero of the function. Let's apply synthetic division with this rational root.
rl IR-0.15cm r - 1/3 & |rr 3 & 1 &-36 &-12

Bring down the first coefficient

rl IR-0.15cm r - 1/3 & |rr 3 & 1 &-36 &-12 &&& & c 3 & & &

Multiply the coefficient by the divisor

rl IR-0.15cm r - 1/3 & |rr 3 & 1 &-36 &-12 &-1 && & c 3 & & &

Add down

rl IR-0.15cm r - 1/3 & |rr 3 & 1 &-36 &-12 &-1 && & c 3 & 0 & &
â–Ľ
Repeat the process for all the coefficients

Multiply the coefficient by the divisor

rl IR-0.15cm r - 1/3 & |rr 3 & 1 &-36 &-12 &-1 & 0 & & c 3 & 0 & &

Add down

rl IR-0.15cm r - 1/3 & |rr 3 & 1 &-36 &-12 &-1 & 0 & & c 3 & 0 &-36 &

Multiply the coefficient by the divisor

rl IR-0.15cm r - 1/3 & |rr 3 & 1 &-36 &-12 &-1 & 0 & 12 & c 3 & 0 & -36 &

Add down

rl IR-0.15cm r - 1/3 & |rr 3 & 1 &-36 &-12 &-1 & 0 & 12 & c 3 & 0 & -36 & 0
Removing the second root, - 13, gave us a quadratic function. 3x^2-36

Factoring the Remaining Quadratic Factor

We will use the Quadratic Formula to find the remaining factors. To do so, we will need to identify the values of a, b, and c. 3x^2+ 0x+( -36)=0 We can see above that a= 3, b= 0, and c= -36. Finally, we will substitute these values into the Quadratic Formula.
x=- b±sqrt(b^2-4ac)/2a
x=- 0±sqrt(0^2-4( 3)( -36))/2( 3)
â–Ľ
Solve for x and Simplify
x=0±sqrt(0-4(3)(-36))/2(3)
x=0±sqrt(0-12(-36))/6
x=0±sqrt(0+432)/6
x=0±sqrt(432)/6
x=± 12sqrt(3)/6
x=± 2sqrt(3)
We can simplify this result into two separate roots 2sqrt(3) and - 2sqrt(3). Finally, let's list all of the roots we have found for the given function. 1, - 1/3, 2sqrt(3), and - 2sqrt(3)