Question

please answer questions no 19

Answer 1

f(x)=ax⁴+2x³-3x²+bx-4
If (x+2) and (x-2) are factors of f(x), then f(-2) and f(2)=0

f(-2) = a(-2)⁴+2(-2)³-3(-2)²+b(-2)-4
16a-16-12-2b-4 = 0
16a-2b-32 = 0

f(2) = a(2)⁴+2(2)³-3(2)²+b(2)-4
16a+16-12+2b-4 = 0
16a+2b=0

16a-2b-32=16a+2b
16a-16a-2b-2b=32
-4b=32
b=32/-4
b=-8
16a+2b=0
16a+2(-8)=0
16a-16=0
16a=16
a=16/16
a=1

a=1
b=-8

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Answer 2

Hi friend,

(x-2) and (x+2) are the zeroes of the given polynomial ax⁴+2x³-3x²+bx-4.

x-2 = 0 ; x = 2

x+2 = 0 ; x = -2

Put x = ±2 to find the values of a and b,

First put x = 2

a(2)⁴+2(2)³-3(2)²+b(2)-4 = 0

16a+2(8)-3(4)+2b-4 = 0

16a + 16 -12 + 2b - 4 = 0

16a + 2b + 4 - 4 = 0

16a + 2b = 0

2(8a+b) = 0

8a + b = 0 --------(1)

Put x = -2,

a(-2)⁴+2(-2)³-3(-2)²+b(-2)-4 = 0

16a+2(-8)-3(4)-2b-4 = 0

16a-16-12-2b-4 = 0

16a-28-2b-4 = 0

16a-32-2b = 0

2(8a-16-b) = 0

8a-b = 16 -------(2)

(1) + (2)

8a + b = 0
8a - b = 16
---------------

16a = 16

a = 16/16 = 1

8a+b = 0

8(1)+b = 0

b = -8

Therefore, a = 1 and b = -8

Hope it helps ……