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Chapter 4: Q. 58 (page 178)

In Problems 57鈥� 62, find the real zeros of f. If necessary, round to two decimal places.

f(x)=x3+3.2x2-7.25x-6.3

Short Answer

Expert verified

The real zeros offafter rounding it two decimal places are2,-4.5,and-0.7.

Step by step solution

01

Step 1. Given Information

The given function isf(x)=x3+3.2x2-7.25x-6.3.

We have to find the real zeros and round them to two decimal places.

02

Step 2. Finding the real zeros 

The degree of the polynomial is 3,thus it has three roots.

Since there are noninteger coefficients, the Rational Zeros Theorem doesn't apply.

03

Step 3. Determine the bounds on the zeros of f 

The leading coefficient of f is1 witha2=3.2,a1=-7.25,anda0=-6.3.

04

Step 4. Evaluate the expressions using the formula 

So,

Max,{1,3.2+-7.25+-6.3}=Max{1,16.75}=16.751+Max{3.2,-7.25,-6.3}=1+7.25=8.25

The smaller of two numbers, 16.75,is the bound.

Every real zero off liesbetween-16.75and16.75.

05

Step 5. Sketch the graph of the polynomial 

From the graph, we conclude that fappears to have three x-intercepts: one between0and-1, one between-4and-5, and one near2.

06

Step 6. Use the Factor Theorem 

f(2)=23+3.2(2)2-7.25(2)-6.3f(2)=8+12.8-14.5-6.3f(2)=0

Therefore, 2is the zero of the polynomial.

Using Zero we find that the remaining zeros are-4.5and-0.7,rounded to two decimal places.

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