Tuesday, July 23, 2019

College Algebra, Chapter 4, 4.4, Section 4.4, Problem 60

The polynomial P(x)=x4+10x2+8x8.

a.) Find all the real zeros of P

The leading coefficient of P is 1, so all rational zeros are integers. They are divisors of constant term 8. Thus, the possible zeros are

±1,±2,±4,±8

Using Synthetic Division







We find that 1,2,4 and 1 are not zeros but that 2 is zero and that P factors as

x4+10x2+8x8=(x+2)(x3+2x2+6x4)

We now factor the quotient x3+2x2+6x4. Its possible zeros are

±1,±2,±4

Using Synthetic Division







We find that 2 is a zero and that P factors as

x4+10x2+8x8=(x+2)(x+2)(x2+4x2)

We now factor the quotient x2+4x2 using Quadratic Formula, we get


x=b±b24ac2ax=4±(4)24(1)(2)2(1)x=2±2


The zeros of P are 2,2+2 and 22.

b.) Sketch the graph of P

No comments:

Post a Comment