Question

3x³-ax²+5x-13 and (a+1)x²-7x+5 leave the same remainder when divided by (x-3)​

Answer 1

Answer:

The value of a is 5.

Step-by-step explanation:

f(x) = 3x³ - ax² + 5x - 13

g(x) = (a + 1)x² - 7x + 5

leave same remainder when divided by (x - 3).

f(3) = 3(3)³ - a(3)² + 5(3) - 13

⇒ 3(27) - a(9) + 15 - 13

⇒ 81 - 9a + 2

∴ 83 - 9a

g(3) = (a + 1)(3)² - 7(3) + 5

⇒ (a + 1)9 - 21 + 5

⇒ 9a + 9 - 21 + 5

⇒ 9a + 9 - 16

∴ 9a - 7

Now, both the remainders are equal.

83 - 9a = 9a - 7

⇒ 83 + 7 = 9a + 9a

⇒ 90 = 18a

∴ a = 5

Answer 2

Given two equations be

$h(x) = 3x^3-ax^2+5x-13 ----- > (1)\\p(x) = (a+1)x^2-7x+5 ----- > (2)$

Given the remainder will be equal when both the equations are divided with $(x-3)$

It means if we substitute $x=3$ these equations will give the remainders of them.

Now,

substituting $x =3$ in equation ($1$) we get

$h(3) = 3(3^3)-a(3^2)+5(3)-13\\h(3) = 83-9a ------- > (3)$

by substituting $x= 3$ in equation ($2$) we get

$p(3) = (a+1)(3^2)-7(3)+5\\p(3) = 9(a+1)-16\\p(3) = 9a-7 ------- > (4)$

now from question we know that both remainders $h(3) \& p(3)$ are equal, then

$h(3)=p(3)\\83-9a=9a-7\\18a=90\\a = 5$

∴The value of $a = 5$

#SPJ2