Question

the polynomials ax^3+3x^2-13 and 2x^3-5x+a are divided by x-2, the remainder in each case is the same. find a​

Answer 1

Sᴏʟᴜᴛɪᴏɴ :-

f(x) = ax³ + 3x² - 13

g(x) = (x - 2)

→ f(g(x)) = a(2)³ + 3(2)² - 13

→ f(2) = 8a + 12 - 13

→ f(2) = 8a - 1

Similarly,

→ f(x) = 2x³ - 5x + a

g(x) = (x - 2)

→ f(g(x)) = 2(2)³ - 5*2 + a

→ f(2) = 16 - 10 + a

→ f(2) = 6 + a .

Now, we have given that, remainder in each case is same .

So,

→ 8a - 1 = 6 + a

→ 8a - a = 6 + 1

→ 7a = 7

→ a = 1 . (Ans.)

Hence, value of a will be 1.

Answer 2

Solution :

The polynomials ax³ + 3x² - 13 & 2x³ - 5x + a are both divided by x-2.

⇒ x-2 = 0

⇒ x = 2

$\underline{\boldsymbol{According\:to\:the\:question\::}}}$

$\longrightarrow\sf{f(x) = g(x) }\\\\\longrightarrow\sf{f(2)=g(2)}\\\\\longrightarrow\sf{a(2)^{3} +3(2)^{2} -13=2(2)^{3} -5(2)+a}\\\\\longrightarrow\sf{a\times 8+3\times 4-13=2\times 8-10+a}\\\\\longrightarrow\sf{8a+12-13=16-10+a}\\\\\longrightarrow\sf{8a-1=6+a}\\\\\longrightarrow\sf{8a-a=6+1}\\\\\longrightarrow\sf{7a=7}\\\\\longrightarrow\sf{a=\cancel{7/7}}\\\\\longrightarrow\bf{a=1}$

Thus;

The value of a will be 1 .