activity 32:

11
ACTIVITY 32: Real Zeros of Polynomials (Section 4.3, pp. 333-340)

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Activity 32:. Real Zeros of Polynomials (Section 4.3, pp. 333-340). Rational Zeros Theorem:. If the polynomial has integer coefficients, then every rational zero of P(x) is of the form where p is a factor of the constant coefficient a 0 and q is a factor of the leading coefficient a n. - PowerPoint PPT Presentation

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ACTIVITY 32:

Real Zeros of Polynomials (Section 4.3, pp. 333-340)

Rational Zeros Theorem:

If the polynomial

has integer coefficients, then every rational zero of P(x) is of the form

where p is a factor of the constant coefficient a0 and q is a factor of the leading coefficient an.

,...)( 011

1 axaxaxaxP nn

nn

,q

p

Example 1:

List all possible rational zeros of P(x) = 2x4 − x2 − 7.

q

p ,1 7,1 2

,1possible rational zeros:

2 of divisors

7 of divisors

,2

1 ,7

2

7

Example 2:

Find the real zeros of f(x) = 2x3 − 5x2 − 4x + 3. Write f(x) in factored form and sketch its graph.

q

p ,1,1 2

,1possible rational zeros:

2 of divisors

3 of divisors

,2

1 ,3

2

3

3

21 5 4

2

2

3

3

7

3

7

4

3 4x - 5x - 2x f(x) 23 ,1possible rational zeros: ,2

1 ,3

2

3

21 5 4

2

2

7

7

3

3

3

0

So -1 is a root which means that

factora is )1(x

)(xQ 13723 452 223 xxxx xx

Consequently, we need only factor

372 2 xx 312 xx

13123 452 23 xxxx xx

13123 452 23 xxxx xx

:roots

1x

2

1x

3x 3,0

Example 3:

Find the real solutions of the equation

06762 234 xxxx

q

p ,1 ,3

1,2

,1possible rational zeros:

1 of divisors

6 of divisors

,2 ,3 6

6

,1possible rational zeros: ,2 ,3 606762 234 xxxx

11 2 6

1

1

3

3

3

7

3

10

6

10

4

11 2 6

1

1

1

1

7

7

7

0

6

0

6

12 2 6

1

2

4

8

2

7

43

6

6

0

possible rational zeros: ,2 ,3 606762 234 xxxx

6762 234 xxxx 324 23 xxx )2(x

2x 324 23 xxx

possible rational zeros: 3324 23 xxx

,1

possible rational zeros: 3324 23 xxx

13 4 2

1

3

7

21

23

3

69

6613 4 2

1

3

1

3

1

3

3

0

6762 234 xxxx 2x 324 23 xxx 3x 2x 12 xx

1a1b1c a

acbbx

2

4)()( 2

1*2

)1(*1*4)1()1( 2

2

411

2

51

2316762 2234 xxxxxxxxConsequently, the real roots of

are x = 3, x = – 2 and

Now we need to find the roots of, x2 – x – 1

2

51x