math question!?

Question

x^4 + 3x^3 - 11x^2 -3x + 10. Solve for all values of x. Show work.

x^4 + 3x^3 - 11x^2 -3x + 10 = 0

Answer 1

search to divide. of10:+/-1;+/-2;+/-5;+/-10.
the answer is:1;-1;2;-5

Answer 2

For one thing, it's not an equation, but I'm going to assume you're equating it to 0. Otherwise, solving for x is an impossible task.

x^4 + 3x^3 - 11x^2 - 3x + 10 = 0

At this point, all we can do is test "plus or minus" factors of 10. That is, 1, -1, 2, -2, 5, -5, 10, -10. We're testing for what makes p(x) equal to 0, because if we get a value of 0, we get a factor.

Let p(x) = x^4 + 3x^3 - 11x^2 - 3x + 10

Test p(1): p(1) = 1 + 3 - 11 - 3 + 10 = 0

Therefore, since 1 is a root, (x - 1) is a factor.

At this point, you have to do long division. Long division on Yahoo answers is extremely difficult, so I'll just give you the answer of what it works out to. You'd basically divide (x - 1) into (x^4 + 3x^3 - 11x^2 - 3x + 10).

The answer works out to x^3 + 4x^2 - 7x - 10 with a remainder of 0. (NOTE: It is absolutely certain that there be a remainder of 0, since we solved that x - 1 is indeed a factor)

So now our factoring brings us this:

(x - 1) (x^3 + 4x^2 - 7x - 10) = 0

At this point, we repeat the process, with
p(x) = x^3 + 4x^2 - 7x - 10, i.e.
we would test p(1), p(-1), p(2), p(-2), p(5), p(-5), p(10), p(-10)

p(1) = 1 + 4 - 7 - 10 = [something non-zero]
p(-1) = -1 + 4 + 7 - 10 = 0

Therefore, since -1 is a root, (x - (-1)), or (x + 1) is a factor.

At this point, again, we do long division:
(x + 1) INTO (x^3 + 4x^2 - 7x - 10).

I can't easily type this out, so I'll solve it on paper and write the answer down. It's up to you to do the long division.

The result of the long division is x^2 + 3x - 10. Thus, so far we have

(x - 1) (x + 1) (x^2 + 3x - 10) = 0

Now, we can factor the quadratic really easily, as we've dealt with a lot of quadratics. x^2 + 3x - 10 factors into (x + 5) (x - 2).

So we get

(x - 1) (x + 1) (x + 5) (x - 2) = 0

At which point, we equate them all to 0, giving us
x = {1, -1, -5, 2}

Answer 3

Let f(x) = x^4 + 3x³ - 11x² -3x + 10

f(1) = 0 so x - 1 is a factor

Thus f(x) = (x - 1)(x³ + 4x² - 7x - 10)

f(2) = 0 so x - 2 is a factor

Thus f(x) = (x - 1)(x - 2)(x² + 6x + 5)
= (x - 1)(x - 2)(x + 1)(x + 5)

Thus x^4 + 3x³ - 11x² -3x + 10 = 0 when x = -5, -1, 1, 2

Answer 4

Assuming it is equal to 0 ..... Let us factorise the LHS
x^4 + 3x^3 - 11x^2 - 3x + 10 = 0
x^4 + 3x^3 - x^2 - 10x^2 - 3x + 10 = 0
x^4 - x^2 + 3x^3 - 3x - 10x^2 + 10 = 0
x^2 (x^2 -1) + 3x (x^2-1) - 10 (x^2-1) = 0
(x^2-1) (x^2 + 3x - 10) = 0
(x-1) (x+1) (x^2+3x-10)=0
(x-1) (x+1) (x+5) (x-2) = 0
So, x = 1, -1, -5, or 2

Answer 5

x^4 +3x^3 - 11x^2 -3x +10=0
find a # that when substituted makes the function above o, so you can make an equasion
x = 1
x-1/ X^4 +3x^3 - 11x^2 -3x +10\ x^3 +4x^2 -7x -10
x^4 - x^3 -
4x^3 -11X^2
4x^3 - 4x^ 2 -
-7x^2 -3x
-7x^2+7x -
-10x +10
-10x +10 -
r-0

x^4 + 3x^3 -11x^2 -3x +10=0
(x-1)*(x^3 +4x^2 -7x -10)=0
then surge for a# again
x=2
x-2/x^3 + 4x^2 -7x -10\ x^2+ 6x +5
x^3 - 2x^2 -
6x^2 - 7x
6x^2 -12x -
5x-10
5x-10 -
r-0
(x-1)(x-2)(x^2 +6x +5)=0
(x-1)(x-2)(x+1)(x+5)=0
x=1or x= 2or x= -1or x=-5
the selution ={-5,-1,1,2}

Answer 6

(x-2)(x-1)(x+1)(x+5)

so the zeroes of x are: 2, 1, -1, -5

Answer 7

You won't learn if you don't do it yourself...