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.

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.

Step-by-step 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

Both of these are equal to R.

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

Multiply both sides by 4:

Distribute:

Subtract 4m on both sides:

Add 47 on both sides:

Simplify left hand side:

Divide both sides by 5:

So the value for m is 7.

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:

Let x=-r/q

This is equal to 4 so we have this equation:

Subtract 4 on both sides:

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:

Compare this to

Since the coefficient of is not 1, we have to find two numbers that multiply to be and add up to be .

Those numbers are 6 and 3 because while .

So we are going to replace or with then factor by grouping:

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:

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).