Question

find the value of a and b so that 3x⁴- 5x³ - 6x² + ax +b is exactly divisible by x² -4​

Answer 1

Step-by-step explanation:

As it is given that

$x {}^{2} - 4$

is exactly divisible by the equation.

x²-4 = ( x-2 )( x+2)

So x-2 = 0 => x=2

x+2=0 => x = -2 are roots of the given equation.

If they are roots by substituting -2 and 2 the equation becomes zero.

therefore substitute 2 and -2 in the equation.

f(x) = 3x⁴ - 5x³ - 6x² + ax + b

f(2) = 3(2)⁴ - 5(2)³- 6(2)² +a(2) + b = 0

48 - 40 - 24 +2a + b =0

2a +b =16 ----------------(1)

now f(-2) = 3(-2)⁴- 5(-2)³ - 6(-2)² + a(-2) + b =0

48+ 40- 24 -2a+b =0

-2a + b = -64 -------------(2)

solve 1 and 2 to get a and b values

2a + b = 16

-2a+ b= -64

-----------------

2b = 16-64

------------------

2b = -48

b= -24

put b in 1 or 2

2a + b = 16

2a -24 = 16

2a = 16 +24

2a = 40

a = 20

Therefore a and b are 20 , -24

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