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Core Connections Algebra 2, 2013

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Core Connections Algebra 2, 2013 View details

Chapter Closure

Exercise 144 Page 656

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

x=1, -2, ±sqrt(2)i

We want to find the zeros of the given polynomial function. To do so, we need to solve the equation f(x)=0. x^4+x^3+2x-4=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 - 4, the possible integer roots are ± 1, ± 2, and ± 4. Let's check.

x	x^4+x^3+2x-4	f(x)=x^4+x^3+2x-4
1	1^4+ 1^3+2( 1)-4	0 ✓
- 1	( -1)^4+( -1)^3+2( -1)-4	-6 *
2	2^4+ 2^3+2( 2)-4	24 *
- 2	( -2)^4+( -2)^3+2( -2)-4	0 ✓
4	4^4+ 4^3+2( 4)-4	324 *
- 4	( -4)^4+( -4)^3+2( -4)-4	180 *

We found that 1 and - 2 are roots for P(x)=0. Therefore, (x-1) and (x+2) 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 1& 1 & 0 & 2 & -4

Bring down the first coefficient

rl IR-0.15cm r 1 & |rr 1& 1 & 0 & 2 & -4 & & & & & c 1& & & &

Multiply the coefficient by the divisor

rl IR-0.15cm r 1 & |rr 1 & 1 & 0 & 2 & -4 & 1 & & & & c 1& & & &

Add down

rl IR-0.15cm r 1 & |rr 1 & 1 & 0 & 2 & -4 & 1 & & & & c 1& 2 & & &

▼

Repeat the process for all the coefficients

Multiply the coefficient by the divisor

rl IR-0.15cm r 1 & |rr 1 & 1 & 0 & 2 & -4 & 1 & 2 & & & c 1& 2 & & &

Add down

rl IR-0.15cm r 1 & |rr 1 & 1 & 0 & 2 & -4 & 1 & 2& & & c 1& 2 & 2 & &

Multiply the coefficient by the divisor

rl IR-0.15cm r 1 & |rr 1 & 1 & 0 & 2 & -4 & 1 & 2 & 2 & & c 1& 2 & 2 & &

Add down

rl IR-0.15cm r 1 & |rr 1 & 1 & 0 & 2 & -4 & 1 & 2& 2 & & c 1& 2 & 2 & 4 &

Multiply the coefficient by the divisor

rl IR-0.15cm r 1 & |rr 1 & 1 & 0 & 2 & -4 & 1 & 2 & 2 &4 & c 1& 2 & 2 & 4 &

Add down

rl IR-0.15cm r 1 & |rr 1 & 1 & 0 & 2 & -4 & 1 & 2& 2 &4 & c 1& 2 & 2 & 4 & 0

Using synthetic division to remove the first root, -1, left us with the following polynomial. x^3+2x^2+2x+4 Following the same procedure, we can factor out (x+2) from 2x^3+2x^2+2x+4.

rl IR-0.15cm r -2 & |rr 1& 2 & 2 & 4

▼

Perform synthetic division

Bring down the first coefficient

rl IR-0.15cm r -2 & |rr 1& 2 & 2 & 4 & & & & c 1& & &

Multiply the coefficient by the divisor

rl IR-0.15cm r -2 & |rr 1& 2 & 2 & 4 & -2 & & & c 1& & &

Add down

rl IR-0.15cm r -2 & |rr 1& 2 & 2 & 4 & -2 & & & c 1& 0& &

Multiply the coefficient by the divisor

rl IR-0.15cm r -2 & |rr 1& 2 & 2 & 4 & -2 & 0& & c 1& 0& &

Add down

rl IR-0.15cm r -2 & |rr 1& 2 & 2 & 4 & -2 & 0& & c 1& 0& 2 &

Multiply the coefficient by the divisor

rl IR-0.15cm r -2 & |rr 1& 2 & 2 & 4 & -2 & 0& -4 & c 1& 0& 2 &

Add down

rl IR-0.15cm r -2 & |rr 1& 2 & 2 & 4 & -2 & 0& -4 & c 1& 0& 2 & 0

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

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. 1x^2+ 0x+ 2=0 We can see above that a= 1, b= 0, and c= 2. Now, let's substitute these values into the Quadratic Formula to find the remaining roots.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- 0±sqrt(0^2-4( 1)( 2))/2( 1)

▼

Solve for x and Simplify

Identity Property of Addition

x=± sqrt(0^2-4(1)(2))/2(1)

Calculate power

x=± sqrt(0-4(1)(2))/2(1)

Multiply

x=± sqrt(0-8)/2(1)

Subtract term

x=± sqrt(-8)/2(1)

SplitIntoFactors

Split into factors

x=± sqrt(4(2)(-1))/2(1)

sqrt(a* b)=sqrt(a)*sqrt(b)

x=± sqrt(4) (sqrt(2))sqrt(-1)/2

i = sqrt(- 1)

x=± sqrt(4) (sqrt(2)) i/2

Calculate root

x=± 2sqrt(2) i/2

Calculate quotient

x=± sqrt(2)i

Therefore, the two other roots are ± sqrt(2)i. Finally, let's list all of the roots we have found for the given function. x=1, -2, sqrt(2)i,and-sqrt(2)i

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