Respuesta :
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
[tex]P(1)=2(1)^3+m(1)^2-5[/tex]
[tex]=2+m-5=m-3[/tex]
[tex]P(\frac{-3}{2})=2(\frac{-3}{2})^3+m(\frac{-3}{2})^2-5[/tex]
[tex]=2(\frac{-27}{8})+m(\frac{9}{4})-5[/tex]
[tex]=-\frac{27}{4}+\frac{9m}{4}-5[/tex]
[tex]=\frac{-27+9m-20}{4}[/tex]
[tex]=\frac{9m-47}{4}[/tex]
Both of these are equal to R.
[tex]m-3=R[/tex]
[tex]\frac{9m-47}{4}=R[/tex]
I'm going to substitute second R which is (9m-47)/4 in place of first R.
[tex]m-3=\frac{9m-47}{4}[/tex]
Multiply both sides by 4:
[tex]4(m-3)=9m-47[/tex]
Distribute:
[tex]4m-12=9m-47[/tex]
Subtract 4m on both sides:
[tex]-12=5m-47[/tex]
Add 47 on both sides:
[tex]-12+47=5m[/tex]
Simplify left hand side:
[tex]35=5m[/tex]
Divide both sides by 5:
[tex]\frac{35}{5}=m[/tex]
[tex]7=m[/tex]
So the value for m is 7.
[tex]P(x)=2x^3+7x^2-5[/tex]
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:
[tex]P(\frac{-r}{q})=2(\frac{-r}{q})^3+m(\frac{-r}{q})^2-5[/tex]
Let x=-r/q
This is equal to 4 so we have this equation:
[tex]2u^3+7u^2-5=4[/tex]
Subtract 4 on both sides:
[tex]2u^3+7u^2-9=0[/tex]
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 Â Â | Â 2 Â Â 7 Â Â 0 Â Â -9
   |     2    9    9
    -----------------------
     2   9   9    0
This gives us the quadratic equation to solve:
[tex]2x^2+9x+9=0[/tex]
Compare this to [tex]ax^2+bx+c=0[/tex]
[tex]a=2[/tex]
[tex]b=9[/tex]
[tex]c=9[/tex]
Since the coefficient of [tex]x^2[/tex] is not 1, we have to find two numbers that multiply to be [tex]ac[/tex] and add up to be [tex]b[/tex].
Those numbers are 6 and 3 because [tex]6(3)=18=ac[/tex] while [tex]6+3=9=b[/tex].
So we are going to replace [tex]bx[/tex] or [tex]9x[/tex] with [tex]6x+3x[/tex] then factor by grouping:
[tex]2x^2+6x+3x+9=0[/tex]
[tex](2x^2+6x)+(3x+9)=0[/tex]
[tex]2x(x+3)+3(x+3)=0[/tex]
[tex](x+3)(2x+3)=0[/tex]
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:
[tex]x \in \{-3,\frac{-3}{2},1\}[/tex]
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).
