Advanced Math/pre calc

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Expert: - 4/29/2009


Question
1. How does the graph of f(x)= -x^2+1 compare to that of g(x)= x^2 ?

2. Use long division to divide (x^3+ 7x^2+ 6x- 18) by (x+3)

3. Write the polynomial 2x^4- 16x^2 - 18 in completely factored form.

4. List the possible rational roots for the function whose equation is f(x)=2x^3+bx^2-cx+10 where b and c are intergers.

Answer

Synthetic Division
1. The first is a parabola opening down and centered at x=0.
The second is a parabola opening up and also centered at x=0.
At the center point, the value of the first is 1
whereas the value of the second is 0.

2. The answer to 2 will be added as an attachment.

3. Note that the powers are even,
which makes it like factoring a parabola.
Factor out the 2 first, and you have 2(x^4 - 8x - 9).

Now if you had t² - 8t - 9,
that factors could be seen to be (t-9)(t+1).
In this problem, just think of the x² as the t I just showed.
To factor it gives 2(x²-9)(x²+1).

We all should recognize that 9 as 3²,
which makes x²-9 into (x-3)(x+3).
This makes the final factorization 2(x-3)(x+3)(x²+1).

4. It should be noted that 2 is prime, whereas 10 is 2*5.
The easiest way for me to do it is to make c into a + c,
then multiply c by a negative when done.
Then again, you could just ignore that and look and see all the roots.

The relation of b to 2 and 10 to c could be the same.
That's 2x^3+bx^2+cx+10, with a -1 times c when done.

Solution 1: We could use 2, so 2*2=4=b,
so since 4=2*2, we can see that c=10/2=5.
This would give 2x^3 + 4x² +5x + 10, which is
2x²(x+2) + 5(x+2) = (2x²+5)(x+2); solution x=-2.

Solution 2: We could use -2, so b=-4,
so since -4=2*(-2), so 10/(-2) = -5=c.
This would give 2x^3 - 4x² -5x + 10.
The factors are 2x²(x-2) - 5(x-2) = (2x²-5)(x-2); solution is x=2.
Note that here, 2x²-5 has a root of x = ±√5, but that's not an integer.

Solution 3: Perhaps we could use 1, so b=2, so 2/2=1, so c=10.
2x^3 + 2x^2 + 10x + 10 = 2x²(x+1) + 10(x+1) = 2(x²+5)(x+1).

Solution 4: Perhaps we could use -1, so b=-2, so 2/(-2)=-1, so c=-10.
2x^3 - 2x^3 - 10x + 10 = 2x²(x+1) - 10(x+1) = (2x²-10)(x+1).
Note that here, x=±√5, but that's not an integer.


Work on problems like this and you can see that the roots are all factors of 10.  That is ±1 and ±2, which both go into 10.  
It might also be possible to get ±5 or ±10.
Try it out and see.

Scott A Wilson

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