polynomial x4+7x3+7x2+px+q is exactly divisible by x2+7x+12, then find

1.	polynomial x4+7x3+7x2+px+q is exactly divisible by x2+7x+12, then find the value of p and q
Answer» Factors of x2 + 7x + 12 :x2 + 7x + 12 = 0{tex}\\Rightarrow{/tex}\xa0x2 + 4x + 3x + 12 = 0{tex}\u200b\u200b\\Rightarrow{/tex}\xa0x(x + 4) + 3(x + 4) = 0{tex}\u200b\u200b\\Rightarrow{/tex}(x + 4) (x + 3) = 0{tex}\u200b\u200b\\Rightarrow{/tex}\xa0x = - 4, -3 ...(i)Since p(x) = x4 + 7x3 + 7x2 + px + qIf p(x) is exactly divisible by x2+ 7x + 12, then x = - 4 and x = - 3 are its zeroes. So putting x = - 4 and x = - 3.p(- 4) = (- 4)4 + 7(- 4)3 + 7(- 4)2 + p(- 4) + qbut p(- 4) = 0{tex}\\therefore{/tex}\xa00 = 256 - 448 + 112 - 4p + q0 = - 4p + q - 80{tex}\\Rightarrow{/tex}4p - q = - 80 ...(i)and p(-3) = (-3)4 + 7(-3)3 + 7(-3)2 + p(-3) + qbut p(-3) = 0{tex}\\Rightarrow{/tex}0 = 81-189 + 63 - 3p + q{tex}\\Rightarrow{/tex}0 = -3p + q -45{tex}\\Rightarrow{/tex}3p -q = -45 ..........(ii)On putting the value of p in eq. (i),we get,{tex}4(-35) - q = - 80{/tex}{tex}\\Rightarrow{/tex}{tex}-140 - q = - 80{/tex}{tex}\\Rightarrow\\ {/tex}{tex}-q = 140 - 80{/tex}{tex}\\Rightarrow\\ {/tex}\xa0{tex}-q = 60{/tex}{tex}\\therefore{/tex}\xa0{tex}q = -60{/tex}Hence, p = -35, q = -60

polynomial x4+7x3+7x2+px+q is exactly divisible by x2+7x+12, then find the value of p and q