Question

Which of the following represents the zeros of f(x) = 2x3 − 5x2 − 28x + 15?

5, −3, −1 over 2
5, −3, 1 over 2
5, 3, −1 over 2
5, 3, 1 over 2

Answer 1

It is 5, -3, $\frac{1}{2}$
So B
I just took the test and got it correct!

Answer 2

Answer: B) 5, −3, 1 over 2 are the zeros of f(x).

Step-by-step explanation:

Given function f(x)=$2x^3-5x^2-28x+15$

If a is a zero of f(x) then f(a)=0

Let's check all the options

$f(5)=2(5)^3-5(5)^2-28(5)+15\\=2(125)-5(25)-28(5)+15=250-125-140+15=0$

$f(-3)=2(-3)^3-5(-3)^2-28(-3)+15\\=2(-27)-5(9)-28(-3)+15=-54-45+84+15=0$

$f(\frac{-1}{2})=2\frac{-1}{2}()^3-5(\frac{-1}{2})^2-28(\frac{-1}{2})+15\\=2(\frac{-1}{8})-5(\frac{-1}{4})+14+15=\frac{-1}{4}-\frac{5}{4}+29=\frac{110}{4}=\frac{55}{2}\neq 0$

Therefore , 5 and -3 are zeroes of f(x).but -1/2 is not.

$f(\frac{1}{2})=2\frac{1}{2}()^3-5(\frac{1}{2})^2-28(\frac{1}{2})+15\\=2(\frac{1}{8})-5(\frac{1}{4})-14+15=\frac{1}{4}-\frac{5}{4}-14+15=0$

⇒ 1/2 isthe third zero of f(x).

Here we got our three zeroes as 5,-3,1/2,rest other will not satisfy f(a)=0

Therefore B represents the zeroes of f(x).