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

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!

total of apples - take away four

6-4 = 2

answer is 2 apples