2 years ago

Question 8 of 10

Mathematics

1 answer:

Alex787 [66]2 years ago

4 0

The answer is c hope I helped!

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Find the rational zeros of the polynomial function, f(x)= 4x^3-8x^2-19x-7

Dima020 [189]

Answer:

The rational zero of the polynomial are $\pm \frac{7}{4}, \pm \frac{1}{4},\pm \frac{7}{2},\pm \frac{1}{2},\pm 7,\pm 1$ .

Step-by-step explanation:

Given polynomial as :

f(x) = 4 x³ - 8 x² - 19 x - 7

Now the ration zero can be find as

$\dfrac{\textrm factor of P}{\textrm factor Q}$ ,

where P is the constant term

And Q is the coefficient of the highest polynomial

So, From given polynomial , P = -7 , Q = 4

Now , $\dfrac{\textrm factor of \pm P}{\textrm factor of \pm Q}$

I.e $\dfrac{\textrm factor of \pm P}{\textrm factor of \pm Q}$ = $\frac{\pm 7 , \pm 1}{\pm 4 ,\pm 2,\pm 1 }$

Or, The rational zero are $\pm \frac{7}{4}, \pm \frac{1}{4},\pm \frac{7}{2},\pm \frac{1}{2},\pm 7,\pm 1$

Hence The rational zero of the polynomial are $\pm \frac{7}{4}, \pm \frac{1}{4},\pm \frac{7}{2},\pm \frac{1}{2},\pm 7,\pm 1$ . Answer

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