Question

the polynomial ax^2 +3x^2 -3 and 2x^2 -5x + a when divided by X-4 leaves the remainder r1 and r2. Find the value of a if r1=r2 is 0

Answer 1

Solution :-

→ f(x) = ax² + 3x² - 3

→ g(x) = (x - 4)

it is given that, when f(x) is divided by g(x) remainder is r1.

So, Putting g(x) Equal to zero.

→ x - 4 = 0

→ x = 4

Putting this value in f(x) , we will get our Remainder .

Thus,

→ f(x) = ax² + 3x² - 3

→ f(4) = a(4)² + 3(4)² - 3

→ f(4) = 16a + 16*3 - 3

→ r1 = 16a + 48 - 3

→ r1 = (16a + 45)

___________________

Similarly,

→ f(x) = 2x² - 5x + a

→ g(x) = (x - 4)

So,

→ f(4) = 2(4)² - 5*4 + a

→ r2 = 2*16 - 20 + a

→ r2 = 32 - 20 + a

→ r2 = (12 + a)

____________________

Now, we have given that, r1 = r2 .

Therefore,

→ 16a + 45 = 12 + a

→ 16a - a = 12 - 45

→ 15a = (-33)

→ a = (-33)/15

→ a = (-11)/5 (Ans.)

Hence, value of a will be (-11/5) .