Question

find the values of a and b so that the polynomial (x⁴+ax³-7x²-8x+b)is exactly divisible by (x+2) as well as (x+3)
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Answer 1

Answer:

Use the factor theorem!

Step-by-step explanation:

Let f(x)=x^4+ax^3-7x^2-8x+b.

x+2=0

or, x=-2

x+3=0

or, x=-3.

For f(x) to be a factor of the expressions (x+2) & (x+3), f(-2) & f(-3) must be equal to zero (0) respectively.

f(-2)=(-2)^4+a(-2)^3-7(-2)^2-8(-2)+b=16-8a-28+16+b=-8a+b+4.

*For f(x) to be divisible by (x+2), f(-2)=0, so,

-8a+b+4=0----(1)

f(-3)=(-3)^4+a(-3)^3-7(-3)^2-8(-3)+b=81-27a-63+24+b=-27a+b+42

*For f(x) to be divisible by (x+3), f(-3)=0, so,

-27a+b+42=0----(2)

Subtracting (2) from (1):

19a-38=0

or, 19a=38

or, a=2.

Substituting a=2 in eq.(1)

-8(2)+b+4=0

or, b=12.

Ans: For x^4+ax^3-7x^2-8x+b to be exactly divisible by (x+2) & (x+3), a=2 & b=12.

Hope it helps!  

Answer 2

Step-by-step explanation:

hope it helps you