   Chapter 3.4, Problem 26E ### Precalculus: Mathematics for Calcu...

7th Edition
James Stewart + 2 others
ISBN: 9781305071759

#### Solutions

Chapter
Section ### Precalculus: Mathematics for Calcu...

7th Edition
James Stewart + 2 others
ISBN: 9781305071759
Textbook Problem

# Integer Zeros All the real zeros of the given polynomial are integers. Find the zeros, and write the polynomial in factored form.26. P(x) = x4 − 2x3 − 3x2 + 8x − 4

To determine

All the real zeroes of a given polynomial and express polynomial in factored form.

Explanation

Given:

The polynomial is P(x)=x42x33x2+8x4 .

Calculation:

The leading coefficient is 1 and constant coefficient is 4 , so the possible real zeros of the

polynomial are factors of 4 : ±1 , ±2 , ±4 .

Test each of the possibilities then substitute the values of x in polynomial P(x) .

To test whether 1 is a zero.

P(x)=x42x33x2+8x4

Substitute 1 for x ,

P(1)=(1)42(1)33(1)2+8(1)4=1+2384=12

Since remainder is 12 , 1 is not a zero of polynomial P(x)=x42x33x2+8x4 .

To test whether 1 is a zero.

P(x)=x42x33x2+8x4

Substitute 1 for x ,

P(1)=(1)42(1)33(1)2+8(1)4=123+84=0

Since remainder is 0 , 1 is a zero of polynomial P(x)=x42x33x2+8x4 .

To test whether 2 is a zero.

P(x)=x42x33x2+8x4

Substitute 2 for x ,

P(2)=(2)42(2)33(2)2+8(2)4=16+1612164=0

Since remainder is 0 , 2 is a zero of polynomial P(x)=x42x33x2+8x4 .

To test whether 2 is a zero.

P(x)=x42x33x2+8x4

Substitute 2 for x ,

P(2)=(2)42(2)33(2)2+8(2)4=161612+164=0

Since remainder is 0 , 2 is a zero of polynomial P(x)=x42x33x2+8x4 .

To test whether 4 is a zero.

P(x)=x42x33x2+8x4

Substitute 4 for x ,

P(4)=(4)42(4)33(4)2+8(4)4=256+12848324=300

Since remainder is 300 , 4 is not a zero of polynomial P(x)=x42x33x2+8x4

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