Question

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

Answer 1

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