Question

When bsmiddle" class="latex-formula"> is divided by $(x-1)(x+2)$, the remainder is $(ax+b)$. This result may be expressed as the identity
$2x^3-4x^2-5x-2$≡$(x-1)(x+2)Q(x)+ax+b$

where Q(x) is the quotient.


a) State the degree of Q(x).

b) By substituting suitable values of x, find a and b.

Answer 1

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 $2x^3$.

If we expand (x-1)(x+2) we should see it's leading term is $x(x)=x^2$

Since $\frac{2x^3}{x^2}=2x$ 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:

$2x^3-4x^2-5x-2=(x-1)(x+2)Q(x)+ax+b$

We want to be careful to choose values for x so that the expression for Q(x) doesn't matter.

If x=1, then $(x-1)(x+2)Q(x)$ would be 0 since x-1 is zero at x=1.  

That is plugging in 1 into term gives you:

$(1-1)(1+2)Q(1)$

$(0)(3)Q(1)$

$0Q(1)$

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:

$2(1)^3-4(1)^2-5(1)-2=(1-1)(1+2)Q(1)+a(1)+b$

$2-4-5-2=0Q(1)+a+b$

$-2-7=0+a+b$

$-9=a+b$

Now notice the factor x+2 in $(x-1)(x+2)Q(x)$.

x+2 is 0 when x=-2 since -2+2=0.

So we are going to plug in -2 into both sides:

$2(-2)^3-4(-2)^2-5(-2)-2=(-2-1)(-2+2)Q(-2)+a(-2)+b$

$2(-8)-4(4)+10-2=-3(0)Q(-2)-2a+b$

$-16-16+8=0-2a+b$

$-32+8=-2a+b$

$-24=-2a+b$

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.

Answer 2

Answer:

a=5

b=-14

I'm not sure if this is what ur lookin for but this is supposed to be the answer