Question

The polynomial P(x) = 2x^3 + mx^2-5 leaves the same remainder when divided by (x-1) or (2x + 3). Find the value of m and the remainder.

The polynomial also leaves the same remainder also leaves the same remainder when divided by (qx+r), find
the values of q and r.​

Answer 1

m=7

Remainder =4

If q=1 then r=3 or r=-1.

If q=2 then r=3.

They are probably looking for q=1 and r=3 because the other combinations were used earlier in the problem.

Explanation:

Let's assume the remainders left when doing P divided by (x-1) and P divided by (2x+3) is R.

By remainder theorem we have that:

P(1)=R

P(-3/2)=R

`P(1)=2(1)^3+m(1)^2-5`

`=2+m-5=m-3`

`P((-3)/(2))=2((-3)/(2))^3+m((-3)/(2))^2-5`

`=2((-27)/(8))+m((9)/(4))-5`

`=-(27)/(4)+(9m)/(4)-5`

`=(-27+9m-20)/(4)`

`=(9m-47)/(4)`

Both of these are equal to R.

`m-3=R`

`(9m-47)/(4)=R`

I'm going to substitute second R which is (9m-47)/4 in place of first R.

`m-3=(9m-47)/(4)`

Multiply both sides by 4:

`4(m-3)=9m-47`

Distribute:

`4m-12=9m-47`

Subtract 4m on both sides:

`-12=5m-47`

Add 47 on both sides:

`-12+47=5m`

Simplify left hand side:

`35=5m`

Divide both sides by 5:

`(35)/(5)=m`

`7=m`

So the value for m is 7.

`P(x)=2x^3+7x^2-5`

What is the remainder when dividing P by (x-1) or (2x+3)?

Well recall that we said m-3=R which means r=m-3=7-3=4.

So the remainder is 4 when dividing P by (x-1) or (2x+3).

Now P divided by (qx+r) will also give the same remainder R=4.

So by remainder theorem we have that P(-r/q)=4.

Let's plug this in:

`P((-r)/(q))=2((-r)/(q))^3+m((-r)/(q))^2-5`

Let x=-r/q

This is equal to 4 so we have this equation:

`2u^3+7u^2-5=4`

Subtract 4 on both sides:

`2u^3+7u^2-9=0`

I see one obvious solution of 1.

I seen this because I see 2+7-9 is 0.

u=1 would do that.

Let's see if we can find any other real solutions.

Dividing:

1 | 2 7 0 -9

| 2 9 9

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

2 9 9 0

This gives us the quadratic equation to solve:

`2x^2+9x+9=0`

Compare this to
`ax^2+bx+c=0`

`a=2`

`b=9`

`c=9`

Since the coefficient of
`x^2` is not 1, we have to find two numbers that multiply to be
`ac` and add up to be
`b`.

Those numbers are 6 and 3 because
`6(3)=18=ac` while
`6+3=9=b`.

So we are going to replace
`bx` or
`9x` with
`6x+3x` then factor by grouping:

`2x^2+6x+3x+9=0`

`(2x^2+6x)+(3x+9)=0`

`2x(x+3)+3(x+3)=0`

`(x+3)(2x+3)=0`

This means x+3=0 or 2x+3=0.

We need to solve both of these:

x+3=0

Subtract 3 on both sides:

x=-3

----

2x+3=0

Subtract 3 on both sides:

2x=-3

Divide both sides by 2:

x=-3/2

So the solutions to P(x)=4:

`x \in \{-3,(-3)/(2),1\}`

If x=-3 is a solution then (x+3) is a factor that you can divide P by to get remainder 4.

If x=-3/2 is a solution then (2x+3) is a factor that you can divide P by to get remainder 4.

If x=1 is a solution then (x-1) is a factor that you can divide P by to get remainder 4.

Compare (qx+r) to (x+3); we see one possibility for (q,r)=(1,3).

Compare (qx+r) to (2x+3); we see another possibility is (q,r)=(2,3).

Compare (qx+r) to (x-1); we see another possibility is (q,r)=(1,-1).