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Sphinxa [80]
3 years ago
6

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 rem

ainder.
The polynomial also leaves the same remainder also leaves the same remainder when divided by (qx+r), find
the values of q and r.​
Mathematics
1 answer:
Zigmanuir [339]3 years ago
6 0

Answer:

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

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

=2+m-5=m-3

P(\frac{-3}{2})=2(\frac{-3}{2})^3+m(\frac{-3}{2})^2-5

=2(\frac{-27}{8})+m(\frac{9}{4})-5

=-\frac{27}{4}+\frac{9m}{4}-5

=\frac{-27+9m-20}{4}

=\frac{9m-47}{4}

Both of these are equal to R.

m-3=R

\frac{9m-47}{4}=R

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

m-3=\frac{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:

\frac{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(\frac{-r}{q})=2(\frac{-r}{q})^3+m(\frac{-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,\frac{-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).

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