# 28, find all the rational zeros of the function. h(t) = t^3 + 8t^2 + 13t + 6

Finding All Zeros of a Polynomial Function Using The Rational Zero Theorem
Finding All Zeros of a Polynomial Function Using The Rational Zero Theorem

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Chapter 2, Problem 23

In Exercises 19 – 28, find all the rational zeros of the function.

\$ h(t) = t^3 + 8t^2 + 13t + 6 \$

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Video by Heather Zimmers

01:30

Find all the rational zeros of the function.
\$h(t)=t^{3}+8 t^{2}+13 t+6\$

02:05

In Exercises 19 – 28, find all the rational zeros of the function.
\$ h(x) = x^3 – 9x^2 + 20x – 12 \$

01:37

Using the Rational Zero Test, find the rational zeros of the function.
\$\$h(t)=t^{3}+8 t^{2}+13 t+6\$\$

03:30

In Exercises 19 – 28, find all the rational zeros of the function.
\$ f(x) = 3x^3 – 19x^2 + 33x – 9 \$

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In Exercises 19 – 28, find all the rational zeros of the function.
\$ f(x) = x^3 – 6x^2 + 11x – 6 \$

06:35

In Exercises 19 – 28, find all the rational zeros of the function.
\$ f(x) = 9x^4 – 9x^3 – 58x^2 + 4x + 24 \$

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In Exercises 21 – 24, find the zeros (if any) of the rational function.
\$ g(x) = \dfrac{x^2 – 9}{x + 3} \$

02:58

In Exercises 19 – 28, find all the rational zeros of the function.
\$ p(x) = x^3 – 9x^2 + 27x – 27 \$

Transcript

What we want to do is find all the rational zeros of this function, and I’m going to start by doing the rational zero test. Irrational zero test will give us a list of all the possible rational zeros, and then we can try some of them. So for the numerator is of all the possible rational zeros, we find the factors of the constant, which is six. So all the factors of six would be a plus or minus one plus or minus to plus or minus three and plus or minus six. And then, for the denominators, we take the leading coefficient, which is one, and we find all the factors of that which would just be plus or minus one. So the possible rational zero’s air all the combinations of these numerator is in these denominators, plus or minus 1/1 plus or minus 2/1 plus or minus 3/1 and plus or minus 6/1. So I have eight numbers in that list, so let’s start trying them. I’m going to start with X equals one, or in this case, T equals one on do synthetic division and see if it comes out evenly. So we’re dividing into one tes cubed plus 80 squared plus 13 T plus six. Bring down the first number wil deployed by one. Write it down and add Take this, multiply it by one Write it down and add Take this multiplied by one. Write it down and add Okay, we didn’t get zero as a…

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