Question

a If x – 1 is a factor of the polynomial p(x) = x3 + ax2 + 2b and a + b = 4, then (a) a = 5, b = -1 (b) a = 9, b= -5 (c) a = 7, b = -3 (d) a = 3, b = 1​

Answer 1

Answer:

option b

Step-by-step explanation:

putting x=1 in equation

1+a+2b=0 (equation 1)

a+b =4 ( equation 2 )

subtract both equations and get the answer

Answer 2

Answer:

Required value of a is 9 and value of b is (-5).

So, option (b) is the correct option.

Step-by-step explanation:

Given polynomial $p(x) = {x}^{3} + a {x}^{2} + 2b...(1)$

It is also given that (x-1) is a factor of polynomial (1).

So, we can say that

$x = 1$ is the solution of $p(x) = 0$

We are putting x = 1 in p(x) = 0,

So,

${1}^{3} + a \times {1}^{2} + 2b = 0 \\ 1 + a + 2b = 0 \\ a + 2b = - 1....(2)$

It is also given that

$a + b = 4.....(3)$

We are Subtracting equation (3) from equation (2),

$a + 2b - (a + b) = - 1 - 4 \\ a + 2b - a - b = - 5 \\ 2b - b = - 5 \\ b = - 5$

From equation (3),

$a - 5 = 4 \\ a = 4 + 5 \\ a = 9$

So, required value of a is 9 and value of b is (-5)

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