Math, asked by Anonymous, 9 months ago

Answer the 20 Questions given in the Attacment​

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Answered by Anonymous
14

Answer:

a) -7

Step-by-step explanation:

Given,

(x+2) and (x-2) are factors of ax^4 + 2x^3 - 3x^2 + bx - 4

So by, Remainder Theorem,

Therefore - 2 and 2 are zeros of the polynomial.

Then, by applying the values of x in P(x), we get,

CASE I :

When (- 2) is the zero

P(x) = ax^4 + 2x^3 - 3x^2 + bx - 4

P(-2) = a(-2)^4 + 2(-2)^3 - 3(-2)^2 + b(-2) - 4

=> 16a - 4 - 16 - 2b - 4 = 0

=> 16a - 2b - 32 = 0

=> 16a - 2b = 32 ......eq (1)

CASE II :

When 2 is the zero

P(x) = ax^4 + 2x - 3x^2 + bx - 4

P(2) = a(2)^4 + 2(2)^3 - 3(2)^2 + b(2) - 4

=> 16a + 16 - 3(4) + 2b - 4 = 0

=> 16a + 2b - 16 + 16 = 0

=> 16a + 2b = 0 ........eq (2)

Now, let us solve equation (1) and equation (2), we get, ;

By subtracting equation (1) from equation (2), we get,

=> 16a + 2b - (16a - 2b) = 0 - 32

=> 16a + 2b - 16a + 2b = -32

=> 4b = - 32 => b = -32/4 = -8

=> b = -8

Now, ny applying the value of b = -2, in equation (1) , we get,

=> 16a - 2(-8) = 32

=> 16a + 16 = 32

=> 16a = 32 - 16

=> 16a = 16

=> a = 16/16 = 1

Hence, b = - 8 and a = 1.

Now, a + b = 1 + (-8) = -7

Hence, a + b = -7

NOTE : THE EXPLANATION OF THIS QUESTION IS CLEARLY GIVEN IN THE ATTACHMENT BELOW. AND THE REAL P(x) = ax^4 + 2x^3 - 3x^2 + bx - 4.

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Answered by psRawaldhiya
1

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