# what are the solutions to the equation: Y=x3+3x2-4x-12?

A:2, -2,3

B:-2,2,-3

C:3,4

D:3,4,12

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• gare
Lv 5

Are A-D our only choices? The actual "solutions" are infinite, in that there is no solution. x may equal any number and by assigning a number to x you get a number for Y. Or, conversely, y may equal any number and you must solve for x. There are an inifinite number of numbers we may assign to X or Y. Using Y as the example, if it is with Y=0, then they do not even need the Y it should just read x^3 + 3x^2 -4x -12 = 0, then we may solve for all values of x relative to one specific answer. Then the question makes sense to select from the answers provided because we are merely looking for the solutions or roots which are exactly the x-intercepts of the function.

Example(s): let y = 0 and solve the problem. Then the (b) answer works. However:

Let y=1, let y = -1, let y = 2, etc. None of the answers now work but it is still an equation that will have a solution.

Ok, y=(x^3)+3(x^2)-4x-12. For solutions I take it you mean the roots of this equation, i.e. values of x for which y=0. Since y is a tri-nomial (x^3 being the highest power) we can expect at most 3 unique roots.

Let's regroup the right hand side to see if we can simplify it a bit.

Y=x3-4x+3x2-12

Y=x(x2-4)+3(x2-4)

Y=(x+3)(x2-4)

Y=(x+3)(x+2)(x-2)

so setting y=0 we indeed get three roots for x, as expected, and they are x=-3, x=+2, and x=-2 so the answer is B

We got lucky in that we could simplify this, you may not always be able to do this! Whether you can simplify it or not to easily pick out the roots, you can always graph the function and see where it crosses the y axis, or where y=0. Hope this helps!

I'm assuming that by solution, you mean roots. Then the answer is B. You factor x^3+3x^2-4x-12

=x^2(x+3)-4(x+3)

=(x^2-4)(x+3)

=(x-2)(x+2)(x+3)

Setting this equal to zero gives you the solutions x=2,-2,-3

Factor by (re)grouping.

y=x³ + 3x² - 4x - 12

y= (x³ + 3x²) + (-4x - 12), factor out GCFs from each group

y=x²(x + 3) - 4(x + 3), rewrite/regroup x² & -4 as a binomial factor of the binomial (x+3)

y=(x² - 4)(x + 3) factor (x² - 4) further..

y=(x + 2)(x - 2)(x + 3), set each binomial equal to zero.

(x+2)=0 or (x-2)=0 or (x+3)=0, solve each for x

x = -2 or x = 2 or x = -3

• Erika
Lv 4
5 years ago

?3/(?(9+y^2 ))dy=?4xdx ----- (a million) combine the left edge enable y = 3 sinh(t) dy=3 cosh(t) dt 9+y^2=9+9 sinh^2(t) = 9(a million+sinh^2 (t)) = 9 cosh^2(t) sqrt(9+y^2)=sqrt(cosh^2(t)) = 3 cosh(t) ?3/(?(9+y^2 ))dy = ? 3 [3 cosh(t) dt] / 3cosh(t) = 3 ? dt = 3 t+C = 3 sinh^-a million(y/3) ?4xdx = 2x^2 (a million) turns into: 3 sinh^-a million(y/3) = 2x^2 + C sinh^-a million(y/3) = (a million/3) {2x^2+C] y/3 = sinh[ (a million/3) (2x^2 + C)] y = 3 sinh[ (a million/3) (2x^2 + C)] it quite is the prevalent answer

• Anonymous

my answer is B:-2,2,-3.....b'coz when substituting these 3 values in the equation we get Y=0..

the solutions are:

-3, -2, 2

Try to use the link below in solving any math problems.

i got it from WAL.

by the way thanks WAL. I learned something from you.