Twice a number, added to seven, is the same
1 answer:
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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
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).