Answer:
a=5
b=-14
I'm not sure if this is what ur lookin for but this is supposed to be the answer
Answer:
a) Q has degree 1.
b) a=5 while b=-14 ( I did this doing the way the problem suggested and I also did the problem using long division.)
Step-by-step explanation:
a) Answer: Q has degree 1.
The left hand side's leading term is [tex]2x^3[/tex].
If we expand (x-1)(x+2) we should see it's leading term is [tex]x(x)=x^2[/tex]
Since [tex]\frac{2x^3}{x^2}=2x[/tex] and 2x has degree 1, then our quotient must have degree 1.
Notes: I multiplied (x-1)(x+2) out in part b if you want to see what it looks like totally expanded.
b. Answer: a=5 while b=-14
You are given the following identity:
[tex]2x^3-4x^2-5x-2=(x-1)(x+2)Q(x)+ax+b[/tex]
We want to be careful to choose values for x so that the expression for Q(x) doesn't matter.
If x=1, then [tex](x-1)(x+2)Q(x)[/tex] would be 0 since x-1 is zero at x=1.
That is plugging in 1 into term gives you:
[tex](1-1)(1+2)Q(1)[/tex]
[tex](0)(3)Q(1)[/tex]
[tex]0Q(1)[/tex]
See what I mean the expression for Q doesn't matter because this result is 0 anyhow.
So let's plug in 1 into both sides:
[tex]2(1)^3-4(1)^2-5(1)-2=(1-1)(1+2)Q(1)+a(1)+b[/tex]
[tex]2-4-5-2=0Q(1)+a+b[/tex]
[tex]-2-7=0+a+b[/tex]
[tex]-9=a+b[/tex]
Now notice the factor x+2 in [tex](x-1)(x+2)Q(x)[/tex].
x+2 is 0 when x=-2 since -2+2=0.
So we are going to plug in -2 into both sides:
[tex]2(-2)^3-4(-2)^2-5(-2)-2=(-2-1)(-2+2)Q(-2)+a(-2)+b[/tex]
[tex]2(-8)-4(4)+10-2=-3(0)Q(-2)-2a+b[/tex]
[tex]-16-16+8=0-2a+b[/tex]
[tex]-32+8=-2a+b[/tex]
[tex]-24=-2a+b[/tex]
So the system to solve is:
a+b=-9
-2a+b=-24
--------------------Subtract the equations to eliminate b:
3a+0=15
3a =15
Divide both sides by 3:
3a/3=15/3
Simplify both sides:
1a=5
a=5
Using one of the equations we found along with a=5 we can not find b.
a+b=-9 with a=5
5+b=-9
Subtract 5 on both sides:
b=-9-5
b=-14
So a=5 while b=-14.
So ax+b is 5x-14.
We could do this another way not suggested by your problem but through long division:
First let's multiply (x-1)(x+2) out using foil.
First: x(x)=x^2
Outer: x(2)=2x
Inner: -1(x)=-x
Last: -1(2)=-2
-------------------Combine like terms:
x^2+x-2
Now let's do the division:
2x-6
----------------------------
x^2+x-2| 2x^3-4x^2-5x-2
-(2x^3+2x^2-4x)
-----------------------
-6x^2 - x -2
-(-6x^2-6x+12)
--------------------
5x-14
We see the remainder is 5x-14. This is what we also got doing as your problem suggested using values for x to plug in to find a and b.
Please let me know if something doesn't make sense to you. Have a good day.
