Question

if f (x) =x^3+px+qx+30 is divisible by (x-5) and when it is divided by (x+6) , the remainder is​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

$\huge\colorbox{yellow}{Solution\:-}$

$f(x) = {x}^{3} + {px}^{2} + 9(x) + 30$

$x - 5 = 0$

$x = 5$

Now, putting the value of x in the equation.

$f(5) = {5}^{3} + p {(5)}^{2} + 9(5) + 30$

$= 125 + 25p + 5p + 30$

$= 25p + 5q + 155 = > 25p + 5q = - 155 \: \: \: \: \: \: \: \: \: \: 1)$

If (x + 6) is divided by f(x), then the remainder = -396

$f( - 6) = {( - 6)}^{3} + p {( - 6)}^{3} + q( - 6) + 30 = - 396$

$= > - 216 + 36p - 6q + 30 = - 396$

$= > 36p - 6q = - 396 + 186$

$36p - 6q = - 210 \: \: \: \: \: \: \: \: \: \: 2)$

From 1 & 2

$150p + 30q = - 930$

$180p - 30q = - 1050$

$330p = - 1980$

$p = \frac{ - 1980}{330}$

$p = - 6$

$25( - 6) + 5q = - 155$

$5q = - 155 + 150$

$5q = - 5$

$q = - 1$

Hence, we concluded that

$\boxed{\pink{p = - 6}}$

$\boxed{\pink{q = - 1}}$

$\huge\colorbox{yellow}{Thank\:You}$