Question

If the polynomials $ax^3+3x^2-13$and $2x^3-5x+a$, when divided by $(x-2)$leave the same remainder find the value of a.

Answer 1

Required Knowledge

[Remainder Theorem]

The remainder theorem states when you divide a polynomial by a linear polynomial, the remainder can be found by substituting the zero of a linear polynomial.

Solution

Let $A(x)=ax^3+3x^2-13$ and $B(x)=2x^3-5x+a$.

By remainder theorem, the polynomials have the same value if x=2.

$\implies A(2)=B(2)$

$\implies 8a+12-13=16-10+a$

$\implies 7a=7$

$\implies \boxed{a=1}$

Proof

[Remainder Theorem]

Let's assume we're dividing by a linear polynomial $P(x)=ax+b$. Then, the remainder should be a constant.

$\implies f(x)=P(x)Q(x)+R$ [Division Algorithm]

If we substitute the zero of $P(x)$ we get,

$\implies f(-\dfrac{b}{a} )=R$

Hence, the remainder of the polynomial is $f(-\dfrac{b}{a} )$.

Answer 2

Solution :-

x - 2 = 0

x = 2 + 0

x = 2

$\sf For\;the\;first\;one$

$\sf p(2)= a(2)^3 + 3(2)^2-13$

$\sf p(2) = a(8) + 3(4)-13$

$\sf p(2) = 8a + 12-13$

$\sf p(2)=8a-1$

$\sf For\;the\;second\;one$

$\sf a(2) = 2(2)^3 - 5(2)+a$

$\sf a(2) = 2(8) - 10+a$

$\sf a(2) = 16 - 10+a$

$\sf a(2) = 6+a$

On comparing both

$\sf 6+a=8a-1$

$\sf 6+1=8a-a$

$\sf 7=7a$

$\sf\dfrac{7}{7} =a$

$\sf 1 = a$