Question

If the polynomial az3+4z2+3z-4 and z3-4z+a leaves the same remainder when divided z-3.find a

Answer 1

Let f(z) = az³+4z²+3z-4
zero of the polynomial (z-3) = 3
(since,z-3=0⇒z=3)
So,replacing z by 3,
f(3) = a(3)³+4(3)²+3(3)-4
⇒f(3) = 27a+36+9-4
⇒f(3) = 27a+41
Let g(z) = z³-4z+a
zero of the polynomial (z-3) = 3
(since,z-3=0⇒z=3)
So,replacing z by 3,
g(3) = (3)³-4(3)+a
⇒g(3) = 27-12+a
⇒g(3) = 15+a
Given that the two polynomials leaves same remainder when divided by z-3
So,g(3)=f(3)
⇒15+a = 27a+41
⇒15-41 = 27a - a
⇒-26 = 26a
⇒a = -1.

Answer 2

The value of a is -1.

Step-by-step explanation:

Given:

The given polynomial is $az^{3} +4z^{2}+3z-4$ and $z^{3} -4z+a$.

When $z-3$ is divided, the remainder is the same.

To find:

To find the value of $a$.

Solution:

The given polynomial is $az^{3} +4z^{2}+3z-4=z^{3} -4z+a$

$z-3=0$

$z=3$

Now substitute $z=3$ in the given polynomial.

$a(3)^{3} +4(3)^{2}+3(3)-4=(3)^{3} -4(3)+a$

$27a+4\times 9+9-4=27-12+a$

$27a+36+9-4=27-12+a$

$27a+45-4=15+a$

$27a+41=15+a$

$27a-a=15-41$

$26a=-26$

$a=-1$
Hence, the value of $a$ is $-1$.

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