Question

Let R1 and R2 are the remainders when polynomials x³+2x²-5ax-7 and x³+ax²-12x+6 are divided by (x+1) and (x-2). If R1+R2=6, find a​

Answer 1

Concept Used :-

Remainder Theorem :-

• This theorem states that when a polynomial p(x) of degree greater than or equal to 1, is divided by linear polynomial x - a, then remainder is p(a).

Solution : -

$\rm :\longmapsto\:Let\:p(x) = {x}^{3} + {2x}^{2} - 5ax - 7$

According to statement,

$\sf \: When \: p(x) \: is \: divided \: by \: x + 1, \: remainder \: is \: R_1$

So,

• By Remainder Theorem,

$\rm :\longmapsto\:R_1 = p( - 1)$

$\rm :\longmapsto\:R_1 = {( - 1)}^{3} + 2 {( - 1)}^{2} - 5a( - 1) - 7$

$\rm :\longmapsto\:R_1 = - 1 + 2 + 5a - 7$

$\rm :\longmapsto\:R_1 = 5a - 6 - - (1)$

Again,

$\rm :\longmapsto\:Let\:q(x) = {x}^{3} + {ax}^{2} - 12 x + 6$

According to statement,

$\sf \: When \: q(x) \: is \: divided \: by \: x - 2, \: remainder \: is \: R_2$

So,

• By Remainder Theorem,

$\rm :\longmapsto\:R_2 = q(2)$

$\rm :\longmapsto\:R_2 = {(2)}^{3} + a {(2)}^{2} - 2 \times 12 + 6$

$\rm :\longmapsto\:R_2 = 8 + 4a - 24 + 6$

$\rm :\longmapsto\:R_2 = 4a - 10 - - - (2)$

Now,

According to given condition,

$\rm :\longmapsto\:R_1 + R_2 = 6$

On substituting the values from equation (1) and (2),

$\rm :\longmapsto\:5a - 6 + 4a - 10 = 6$

$\rm :\longmapsto\:9a - 16 = 6$

$\rm :\longmapsto\:9a = 6 + 16$

$\rm :\longmapsto\:9a = 22$

$\bf\implies \:a = \dfrac{22}{9}$

Additional Information :-

Factor Theorem :-

• This theorem states that when a polynomial p(x) of degree greater than or equal to 1, is divided by linear polynomial x - a, then remainder p(a) is 0.