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

Answer:The rational zero of the polynomial are [tex]\pm \frac{7}{4}, \pm \frac{1}{4},\pm \frac{7}{2},\pm \frac{1}{2},\pm 7,\pm 1[/tex]  .  Step-by-step explanation:Given polynomial as :f(x) = 4 x³ - 8 x² - 19 x - 7Now the ration zero can be find as [tex]\dfrac{\textrm factor of P}{\textrm factor Q}[/tex] , where P is the constant termAnd Q is the coefficient of the highest polynomial So, From given polynomial ,  P = -7 , Q = 4Now , [tex]\dfrac{\textrm factor of \pm P}{\textrm factor of \pm Q}[/tex]I.e  [tex]\dfrac{\textrm factor of \pm P}{\textrm factor of \pm Q}[/tex] = [tex]\frac{\pm 7 , \pm 1}{\pm 4 ,\pm 2,\pm 1 }[/tex] Or, The rational zero are [tex]\pm \frac{7}{4}, \pm \frac{1}{4},\pm \frac{7}{2},\pm \frac{1}{2},\pm 7,\pm 1[/tex]Hence The rational zero of the polynomial are [tex]\pm \frac{7}{4}, \pm \frac{1}{4},\pm \frac{7}{2},\pm \frac{1}{2},\pm 7,\pm 1[/tex]  .  Answer