MATH help...?

Question

determine the remainder


(x³-4x²+5x-2) divided by x-1




(6y³+y²+9y+5) divided by (2y+1)


can u please explain how do to them






solve each inequality

x(x-3)>0



x³+4x²-11x-30 > & equal to 0

Answer 1

1) Let p(x) = x^3 - 4x^2 - 5x - 2.
If divided by (x - 1), all that's required to find the remainder is to find
p(1).

p(1) = 1^3 - 4(1)^2 - 5(1) - 2 = 1 - 4 - 5 - 2 = -10

So the remainder is -10.

2) Let p(y) = 6y^3 + y^2 + 9y + 5.
If divided by 2y + 1, it follows that whatever makes 2y + 1 = 0, and plugged into p(y), will give us the remainder. The answer is
y = -1/2, so

p(-1/2) = 6(-1/2)^3 + (-1/2)^2 + 9(-1/2) + 5
p(-1/2) = 6(-1/8) + 1/4 - 9/2 + 5
p(-1/2) = -3/4 + 1/4 - 18/4 + 20/4
p(-1/2) = -2/4 + 2/4 = 0

Therefore, the remainder is 0.

3)

x(x - 3) > 0

To solve this inequality, we first require the critical values. Solve
x(x - 3) = 0. This gives us the solutions x = {0, 3}, so our critical values are 0 and 3.

Now, we want to test a point in these three regions:
a) (-infinity, 0)
b) (0, 3)
c) (3, infinity)

What we want to test for are *positive* numbers, since the inequality specifies "greater than 0."

For a), test -1. Then x(x - 3) = (-1)(-1 -3) = 4, which is positive.
Therefore, keep the interval (-infinity, 0).

For b), test 1. Then x(x - 3) = 1(1 - 3) = -2, which is negative. Reject this interval.

For c), test 10. Then 10(10 - 3) = 70, which is positive.
Keep this interval (3, infinity).

Therefore, the solution set is
x an element of (-infinity, 0) U (3, infinity)

{Note, the "U" means "union"}

4) x^3 + 4x^2 - 11x - 30 >= 0

We need to find a way to factor the cubic.

Let p(x) = x^3 + 4x^2 - 11x - 30

What you want to do is test factors of -30; that is, you want to test x = +/- 1, +/- 2, +/- 3, +/- 5, +/- 6, +/- 10, +/- 15, +/- 30.

Note that you do not need to test ALL of these values; as soon as you find one that works, you can get the others easily. You want to test these values to see if any of them will give you a remainder of 0.

p(-1) = -1 + 4 + 11 - 30 = [nonzero]
p(1) = 1 + 4 - 11 - 30 = [nonzero]
p(2) = 2^3 + 4(2)^2 - 11(2) - 30 = 8 + 16 - 22 - 30 = [nonzero]
p(-2) = (-2)^3 + 4(-2)^2 - 11(-2) - 30 = -8 + 16 + 22 - 30 = 0

Since -2 is a root, it follows that (x - (-2)), or (x + 2), is a factor.

At this point, you use synthetic long division.
(x + 2) into (x^3 + 4x^2 - 11x - 30)

Without showing you the details (as long division is difficult to show on here, you should get a quotient of x^2 + 2x - 15 and a remainder of 0. Therefore, we get a factorization of p(x) now.

p(x) = (x + 2)(x^2 + 2x - 15)
p(x) = (x + 2)(x + 5)(x - 3)

Now that we have a factorization of our cubic, we go back to our original question.

x^3 + 4x^2 - 11x - 30 >= 0

Now, factor as we just have.

(x + 2)(x + 5)(x - 3) >= 0

Critical values: x = {-2, -5, 3}

That means we have four significant regions to test.

a) (-infinity, -5]
b) [-5, -2]
c) [-2, 3]
d) [3, infinity)

{Note: since we're dealing with "greater than _or equal to_", we use square brackets for our interval notation because our solution set would include 0.}

For region a), test -10. Then
(x + 2)(x + 5)(x - 3) = (negative)(negative)(negative) = negative.
Reject this interval.

For region b), test -3. Then
(x + 2)(x + 5)(x - 3) = (negative)(positive)(negative) = positive.
Keep the interval [-5. -2]

For region c), test 0 Then
(x + 2)(x + 5)(x - 3) = (positive)(positive)(negative) = negative.
Reject this interval.

For region d), test 10. Then
(x + 2)(x + 5)(x - 3) = (positive)(positive)(positive) = positive.
Keep this interval [3, infinity)

Therefore, our solution set is

x is an element of [-5, -2] U [3, infinity)

Answer 2

x^3 - 4x^2 + 5x - 2 divided by x - 1
Use synthetic division:

1 -3 2
1|1 -4 5 -2
|____________
1 -3 2 0

The quotient is x^2 -3x + 2 with remainder of zero
The quotient can be factored into (x-2)(x-1)

Use synthetic division on the next division problem also. The answer is:
6x^2 - 2x +13 with a remainder of -3/2

x(x-3) > 0
Several steps:
1. we know that for x(x-3) to be greater than zero neither x nor x-3 can equal zero.
Therefore x must not = zero and x must not be -3

2. if both x and x-3 are greater than zero, then their product must be greater than zero. Therefore, one solution is x > 0 AND x>3.

3. if both x and x-3 are less than zero, then their product must be greater than zero. There fore one solution is x < 0 AND x < 3

4. Combining steps 1-3, we get the following solution:
x > 3 or x < 0

the last one is too much work for me. Basically, you'll need to factor the polynomial (using synthetic division with the possible roots, which are +-1, +-2, +-3,
+-5, +-6, +-10, +-15, +-30), then factor the resultant trinomial. then consider all possibilities for the three factors (two factors are positive and one is negative, two are negative and one is positive, all three are positive). Good luck!

Answer 3

It is very difficult to type long division on these answer pages. To learn long division of polynomials you should visit a website like:

http://www.purplemath.com/modules/polydiv2.htm

just search under "division" and "polynomials" and you will find some helpful sites.

Answer 4

Sorry, i would but its the weekend. There is no school until tommorrow.