Question

X^2+ax+b is divided by (x-1) , the remainder is 15 and when x^2 + bx+ a is divided by (x+1) , the remainder is -1 , then value of a^2+ b^2 is

Answer 1

Answer:

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Answer 2

The value of $a^2+b^2$ is 100.

Step-by-step explanation:

According to Remainder theorem, if a polynomial P(x) is divided by (x-c), then P(c) represents the remainder.

It is given that $x^2+ax+b$ is divided by (x-1) , the remainder is 15

Using Remainder theorem we get

$(1)^2+a(1)+b=15$

$1+a+b=15$

$a+b=14$              .... (1)

When $x^2+bx+ a$ is divided by (x+1) , the remainder is -1.

Using Remainder theorem we get

$(-1)^2+b(-1)+ a=-1$

$1-b+a=-1$

$a-b=-2$            .... (2)

On adding (1) and (2)we get

$2a=12$

$a=6$

Substitute a=6 in equation (1).

$6+b=14$  

$b=14-6$

$b=8$

$a^2+b^2=6^2+8^2=36+64=100$

Therefore, the value of $a^2+b^2$ is 100.

#Learn more

P(x)=2x³-x²+kx-4.

Divide p(x) by x-2 remainder will be 6. Find k

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