avamolish1
Answered

Westonci.ca connects you with experts who provide insightful answers to your questions. Join us today and start learning! Discover a wealth of knowledge from professionals across various disciplines on our user-friendly Q&A platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

Find all the zeros of f(x) help please

Find All The Zeros Of Fx Help Please class=

Sagot :

LammettHash

Since [tex]f(6)=0[/tex], by the remainder theorem this means that x - 6 divides f(x) exactly. This means there are constants a, b, and c such that

[tex]\dfrac{f(x)}{x-6} = ax^2 + bx + c[/tex]

Multiplying both sides by x - 6 gives

[tex]f(x) = (x-6)(ax^2+bx+c) \\\\ 2x^3-19x^2+45x-18 = ax^3 + (b-6a)x^2 + (c-6b)x - 6c[/tex]

Then a, b, c satisfy

[tex]\begin{cases}a=2 \\ b-6a=-19 \\ c-6b=45 \\ -6c=-18\end{cases}[/tex]

and solving this system gives a = 2, b = -7, and c = 3.

So, we have

[tex]2x^3-19x^2+45x-18 = (x - 6) (2x^2 - 7x + 3)[/tex]

The quadratic term can be factored as

[tex]2x^2 - 7x + 3 = (2x - 1)(x - 3)[/tex]

which leaves us with

[tex]f(x) = (x-6)(2x-1)(x-3)[/tex]

so that the zeros of f(x) are 6, 1/2, and 3.

We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.