QuestionList the possible rational zeros and then solve to identify the zeros. $f(x)=2 x^{3}-11 x^{2}+18 x-9$

Studdy Solution

STEP 1

What is this asking? We're on a mission to find all the possible rational zeros of this cubic function, and then figure out which ones are actually zeros! Watch out! Don't forget, just because a number could be a zero doesn't mean it is one.
We've gotta test them out!

STEP 2

1. Possible Rational Zeros
2. Test the Zeros
3. Find Remaining Zeros

STEP 3

Let's kick things off with the Rational Root Theorem!
This theorem tells us that any rational zero of our function will be of the form $p/q$ , where $p$ is a factor of the constant term (that's $-9$ ) and $q$ is a factor of the leading coefficient (which is $2$ ).

STEP 4

So, what are the factors of $-9$ ?
Well, we've got $\pm 1$ , $\pm 3$ , and $\pm 9$ .
And for $2$ , the factors are $\pm 1$ and $\pm 2$ .

STEP 5

Now, let's mix and match!
Dividing each factor of $-9$ by each factor of $2$ gives us our possible rational zeros: $\pm 1$ , $\pm \frac{1}{2}$ , $\pm 3$ , $\pm \frac{3}{2}$ , $\pm 9$ , and $\pm \frac{9}{2}$ .
That's quite a list!

STEP 6

Time to roll up our sleeves and test these potential zeros!
We'll use synthetic division because it's super speedy.
Let's start by testing $x = 1$ .

STEP 7

$\begin{array}{cccc} 1 & 2 & -11 & 18 \\ & 2 & -9 & 9 \\ & 2 & -9 & 9 \\ \end{array}$

STEP 8

Boom! We got a remainder of zero, which means $x = 1$ is a zero!
The quotient is $2x^2 - 9x + 9$ .

STEP 9

Now we've got a quadratic equation $2x^2 - 9x + 9 = 0$ , which we can solve using factoring!

STEP 10

We're looking for two numbers that multiply to $2 \cdot 9 = 18$ and add up to $-9$ .
Those numbers are $-6$ and $-3$ .

STEP 11

Rewriting our quadratic, we get $2x^2 - 6x - 3x + 9 = 0$ .
Factoring by grouping gives us $2x(x - 3) - 3(x - 3) = 0$ , which simplifies to $(2x - 3)(x - 3) = 0$ .

STEP 12

Setting each factor equal to zero, we find $2x - 3 = 0$ gives us $x = \frac{3}{2}$ , and $x - 3 = 0$ gives us $x = 3$ .

STEP 13

We found all three zeros!
They are $x = 1$ , $x = \frac{3}{2}$ , and $x = 3$ .

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