Find the values of a and b given that the polynomial P(x) = x^3+ ax^2 +x+b, is divisible by both x-1 and x + 3.​

a=4

b=-6

Step-by-step explanation:

If P(x) is divisible by x-c, then P(c)=0.

So P(1)=0 implies 1^3+a1^2+1+b=0

and

P(-3)=0 implies (-3)^2+a(-3)^2+-3+b=0

So notice we have a system to solve.

Let's simply it.

First equation:

1^3+a1^2+1+b=0

1+a+1+b=0

2+a+b=0

a+b=-2

Second equation:

(-3)^3+a(-3)^2+-3+b=0

-27+9a-3+b=0

9a+b-30=0

9a+b=30

Let's put our system together:

a+b=-2

9a+b=30

This is setup so if we subtract the equations b will eliminate allowing us to solve for a:

-8a=-32

a=4

If a=4 and a+b=-2, then 4+b=-2 giving us b=-6.

a=4

b=-6

answer: the answer is one of

step-by-step explanation:   we are given to write a proper fraction where the whole is divided into an even number of equal parts less than ten, and an odd number of those parts exist.

an even number less than 10 may be one of 2, 4, 6 or 8. so, the whole may be one of these four evens.

now, an odd number of these parts must exist, so the numerator must be an odd number, one of 1, 3, 5 or 7. it cannot be 9 because then the fraction will become improper.

thus, our fraction may be one of the following