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Big Ideas Math Algebra 2, 2014

BI

Big Ideas Math Algebra 2, 2014 View details

5. Solving Polynomial Equations

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Exercise 46 Page 195

By the Irrational Conjugates Theorem, if a+sqrt(b) is a zero of a polynomial with rational coefficients, then a-sqrt(b) is also a zero, where a and b are rational numbers, with sqrt(b) irrational.

f(x)=x^4+5x^3-33x^2-85x

Practice makes perfect

We want to write a polynomial function f that meets the following requirements.

It is of the least possible degree.
It has rational coefficients.
The leading coefficient is 1.
The zeros are 0, 5, and -5+sqrt(8).

Because -5+sqrt(8) is a zero, by the Irrational Conjugates Theorem -5-sqrt(8) is also a zero. This means we have four zeros, so the least degree is 4. Let's write the factored form of f. f(x)= 1(x- 0)(x- 5)(x-( -5+sqrt(8)))(x-( -5-sqrt(8))) Finally, we will simplify the equation by multiplying the factors.

f(x)=1(x-0)(x-5)(x-(-5+sqrt(8)))(x-(-5-sqrt(8)))

Identity Property of Multiplication

f(x)=(x-0)(x-5)(x-(-5+sqrt(8)))(x-(-5-sqrt(8)))

Identity Property of Addition

f(x)=(x)(x-5)(x-(-5+sqrt(8)))(x-(-5-sqrt(8)))

Distribute - 1

f(x)=(x)(x-5)(x+5-sqrt(8))(x+5+sqrt(8))

Add parentheses

f(x)=(x)(x-5)((x+5)-sqrt(8))((x+5)+sqrt(8))

ExpandDiffSquares

(a+b)(a-b)=a^2-b^2

f(x)=(x)(x-5)((x+5)^2-(sqrt(8))^2)

( sqrt(a) )^2 = a

f(x)=(x)(x-5)((x+5)^2-8)

ExpandNegPerfectSquare

(a-b)^2=a^2-2ab+b^2

f(x)=(x)(x-5)(x^2+10x+25-8)

Subtract term

f(x)=(x)(x-5)(x^2+10x+17)

▼

Multiply

Multiply (x) by (x-5)

f(x)=(x^2-5x)(x^2+10x+17)

Distribute (x^2+10x+17)

f(x)=x^2(x^2+10x+17)-5x(x^2+10x+17)

Distribute x^2

f(x)=x^4+10x^3+17x^2-5x(x^2+10x+17)

Distribute -5x

f(x)=x^4+10x^3+17x^2-5x^3-50x^2-85x

Add and subtract terms

f(x)=x^4+5x^3-33x^2-85x

Subchapter links

Communicate Your Answer p.189

Monitoring Progress p.191-192

Exercises p.194-196