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Find value of a and b, if x^2 - 4 is factor of ax^4 + 2x^3 - 3x^2 + bx - 4.
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Answered by
5
→a=1 and b= -8
→ for further details refer to the attachment.
→ for further details refer to the attachment.
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Answered by
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Given f(x) = ax^4 + 2x^3 - 3x^2 + bx - 4.
Given that g(x) = x^2 - 4 = (x + 2)(x - 2) are the factors of f(x).
When x = 2:
f(2) = a(2)^4 + 2(2)^3 - 3(2)^2 + b(2) - 4
= 16a + 16 - 12 + 2b - 4
= 16a + 2b = 0 ------ (1)
When x = -2:
f(-2) = a(-2)^4 + 2(-2)^3 - 3(-2)^2 + b(-2) - 4
= 16a - 16 - 12 - 2b - 4
= 16a - 2b - 32 = 0
= 16a - 2b = 32 ----- (2)
On solving (1) & (2), we get
16a + 2b = 0
16a - 2b = 32
-------------------
32a = 32
a = 1.
Substitute a = 1 in (2), we get
16a + 2b = 0
16 = -2b
b = -8.
Therefore a = 1, b = -8.
Hope this helps!
Given that g(x) = x^2 - 4 = (x + 2)(x - 2) are the factors of f(x).
When x = 2:
f(2) = a(2)^4 + 2(2)^3 - 3(2)^2 + b(2) - 4
= 16a + 16 - 12 + 2b - 4
= 16a + 2b = 0 ------ (1)
When x = -2:
f(-2) = a(-2)^4 + 2(-2)^3 - 3(-2)^2 + b(-2) - 4
= 16a - 16 - 12 - 2b - 4
= 16a - 2b - 32 = 0
= 16a - 2b = 32 ----- (2)
On solving (1) & (2), we get
16a + 2b = 0
16a - 2b = 32
-------------------
32a = 32
a = 1.
Substitute a = 1 in (2), we get
16a + 2b = 0
16 = -2b
b = -8.
Therefore a = 1, b = -8.
Hope this helps!
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