Exercise 37 Page 323

We want to find the zeros of the polynomial function y=P(x). To do so, we need to solve the equation P(x)=0. 2x^4+3x^3-17x^2-27x-9=0

The degree of P(x) is 4. Thus, by the Fundamental Theorem of Algebra, we know that P(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 P(x) is - 9, the possible integer roots are ± 1, ± 3, and ± 9. Let's check.

x	2x^4+3x^3-17x^2-27x-9	P(x)=2x^4+3x^3-17x^2-27x-9
1	2( 1)^4+3( 1)^3-17( 1)^2-27( 1)-9	- 48 *
- 1	2( - 1)^4+3( - 1)^3-17( - 1)^2-27( - 1)-9	0 ✓
3	2( 3)^4+3( 3)^3-17( 3)^2-27( 3)-9	0 ✓
- 3	2( - 3)^4+3( - 3)^3-17( - 3)^2-27( - 3)-9	0 ✓
9	2( 9)^4+3( 9)^3-17( 9)^2-27( 9)-9	13680 *
- 9	2( - 9)^4+3( - 9)^3-17( - 9)^2-27( - 9)-9	9792 *

We found that - 1, 3, and - 3 are roots for P(x)=0. Therefore, (x+1), (x-3), and (x+3) are factors of the polynomial. Let's use synthetic division to factor out those factors and thus find the fourth root. We will start by factoring out (x+1).

rl IR-0.15cm r - 1 & |rr 2& 3 & - 17 & - 27&- 9

Bring down the first coefficient

rl IR-0.15cm r - 1 & |rr 2& 3 & - 17 & - 27&- 9 & & & & & c 2& & & &

Multiply the coefficient by the divisor

rl IR-0.15cm r - 1 & |rr 2& 3 & - 17 & - 27&- 9 &- 2 & & & & c 2& & & &

Add down

rl IR-0.15cm r - 1 & |rr 2& 3 & - 17 & - 27&- 9 &- 2 & & & & c 2& 1& & &

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Repeat the process for all the coefficients

Multiply the coefficient by the divisor

rl IR-0.15cm r - 1 & |rr 2& 3 & - 17 & - 27&- 9 &- 2 &- 1 & & & c 2& 1& & &

Add down

rl IR-0.15cm r - 1 & |rr 2& 3 & - 17 & - 27&- 9 &- 2 &- 1 & & & c 2& 1 & - 18& &

Multiply the coefficient by the divisor

rl IR-0.15cm r - 1 & |rr 2& 3 & - 17 & - 27&- 9 &- 2 &- 1 & 18& & c 2& 1 & - 18& &

Add down

rl IR-0.15cm r - 1 & |rr 2& 3 & - 17 & - 27&- 9 &- 2 &- 1 & 18& & c 2& 1 & - 18& - 9&

Multiply the coefficient by the divisor

rl IR-0.15cm r - 1 & |rr 2& 3 & - 17 & - 27&- 9 &- 2 &- 1 & 18&9 & c 2& 1 & - 18& - 9&

Add down

rl IR-0.15cm r - 1 & |rr 2& 3 & - 17 & - 27&- 9 &- 2 &- 1 & 18&9 & c 2& 1 & - 18& - 9 &0

Using synthetic division to remove the first root, -1, left us with the following polynomial. 2x^3+x^2-18x-9 Following the same procedure, we can factor out (x-3) from 2x^3+x^2-18x-9.

rl IR-0.15cm r 3 & |rr 2& 1 & - 18 & - 9

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Perform synthetic division

Bring down the first coefficient

rl IR-0.15cm r 3 & |rr 2& 1 & - 18 & - 9 & & & & c 2& & &

Multiply the coefficient by the divisor

rl IR-0.15cm r 3 & |rr 2& 1 & - 18 & - 9 &6 & & & c 2& & &

Add down

rl IR-0.15cm r 3 & |rr 2& 1 & - 18 & - 9 &6 & & & c 2& 7& &

Multiply the coefficient by the divisor

rl IR-0.15cm r 3 & |rr 2& 1 & - 18 & - 9 &6 & 21& & c 2& 7& &

Add down

rl IR-0.15cm r 3 & |rr 2& 1 & - 18 & - 9 &6 & 21& & c 2& 7& 3&

Multiply the coefficient by the divisor

rl IR-0.15cm r 3 & |rr 2& 1 & - 18 & - 9 &6 & 21&9 & c 2& 7& 3&

Add down

rl IR-0.15cm r 3 & |rr 2& 1 & - 18 & - 9 &6 & 21&9 & c 2& 7& 3& 0

Removing the second root, 3, gave us an even smaller polynomial. 2x^2+7x+3

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. 2x^2+ 7x+ 3=0 We can see above that a= 2, b= 7, and c= 3. Finally, we will substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- 7±sqrt(7^2-4( 2)( 3))/2( 2)

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Solve for x and Simplify

CalcPow

Calculate power

x=-7±sqrt(49-4(2)(3))/2(2)

Multiply

x=-7±sqrt(49-24)/4

SubTerm

Subtract term

x=-7±sqrt(25)/4

CalcRoot

Calculate root

x=-7± 5/4

We can simplify this result into two separate roots.

x=-7± 5/4
x_1=-7+ 5/4	x_2=-7-5/4
x_1=-1/2	x_2=-3

We already knew that -3 was a root, but now we have found our fourth root, - 12. Finally, let's list all of the roots we have found for the given function.

x=-1, 3, -3,and-1/2

Hints

Check the answer

Practice exercises

Asses your skill

Integer Roots

Factoring the Remaining Quadratic Factor