Answer the 20 Questions given in the Attacment
Answers
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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