Solve: y^3+3y^2-4y-12=0?

Question

... means equal to zero.

Answer 1

First factor by grouping:

y^3+3y^2-4y-12=0
(y^3 + 3y^2) - (4y + 12) = 0
y^2(y + 3) - 4(y + 3) = 0
(y^2 - 4)(y + 3) = 0

Then factor the difference of squares:
(y + 2)(y - 2)(y + 3) = 0

Then set each factor equal to zero:
y = -2 OR y = 2 OR y = -3

Answer 2

2

Answer 3

2

Answer 4

y³ + 3y² - 4y - 12 = 0. [Factor by grouping]
(y³ + 3y²) + (-4y - 12) = 0. [Factor GCF from each group]
y²(y + 3) - 4(y + 3) = 0. [Combine like groups]
(y² - 4)(y + 3) = 0. [Factor difference of two squares]
(y + 2)(y - 2)(y + 3) = 0. [Use zero-product property. Each factor can be equal to zero.]
y + 2 = 0 or y - 2 = 0 or y + 3 = 0. [Solve each for y]
y = -2 or y = 2 or y = 3.
y = ±2 or y = -3.

Answer 5

y^3 + 3y^2 - 4y - 12 = 0

1) Check the graph of the function for potential real zeros:
http://www.coolmath.com/graphit/


OR

potential zeros can be found by dividing all the factors or "-12" by all the factors of the coefficient of "y^3" which are 1 and -1
So potential roots are (-12, 12, -6, 6, -4, 4, -3, 3, -2, 2, -1, 1)
Since there are so many, just use the graph.

Looks like the graph crossed at y = 2

2)Divide out (y - 2) factor and obtain a quadratic:

2 |___1___3__-4_-12
_________2__10__12
______1__5___6___0 remainder

_1__5___6___0 ----> x^2 + 5x + 6

3)Solve the quadratic:

x^2 + 5x + 6 = 0
(x + 3)(x + 2) = 0

x = -2, -3 and 2

Answer 6

y3-3y2-5y+15 4 time period polynomials are better accessible to element than trinomials ... once you element a trinomial you need to verify on factors that when stronger equivalent your 1st and extremely last time period, and at the same time as further equivalent your center time period. yet, once you element a 4 time period polynomial each and each and every of your 4 words is a seperate fabricated out of your 2 factors... you'll see what I recommend. There are 4 words : y³, - 3y², - 5y, and + 15 y³ can merely be factored one way (assuming merely 2 factors): (y)(y²) ... position y and y² because the first words of your factors. (y ... )(y² ...) Your consistent (15) is effective. that signifies that the signs and indicators on your factors will be both effective or both negative. depending on the second one and 0.33 words on your polynomial, - 3y² and - 5y, you recognize that your signs and indicators will both be negative. (y - ? )(y² - ? ) The fabricated out of your very last 2 factors is 15 ... element 15: (a million,15) (3,5) Is there a fashion to combine this variety of instruments of factors so as that your very last time period is 15 and your center words are - 3y² and - 5y ? (y - 3)(y² - 5) = y³ - 3y² - 5y + 15 y³ - 3y² - 5y + 15 = 0 (y - 3)(y² - 5) = 0 y - 3 = 0 y = 3 y² - 5 = 0 y² = 5 y = ?5 Your zeroes are: 3, ?5 P.S. once you spot that the constants on your 2 center words (- 3)(- 5) are factors of your consistent (+15) you could commence to suspect that your polynomial will element actual. i wish it really is sensible, sturdy success.

Answer 7

y = 2 (by trial and error) is a solution but we know that the equation has three solutions since highest exponent is 3.

now use long division...that is to say... divide your equation by (y-2) and we get...

(y-2)(y^2 + 5y +6) = (y-2)(y+3)(y+2) = y^3 + 3y^2 + - 4y - 12 = 0.

So, solutions are y = 2, -3, and -2.

Answer 8

y^3+3y^2-4y-12=0
I see y=2 is a solution. So y-2 is a factor of y^3+3y^2-4y-12.
y^3+3y^2-4y-12 = (y-2)(y2 + 5y + 6) = (y-2)((+2)(y+3)

y^3+3y^2-4y-12=0 if (y-2)((y+2)(y+3) = 0
That is true if y=2, y=-2 or if y = -3
Th

Answer 9

y^3 + 3y^2 - 4y - 12 = 0
(y^3 + 3y^2) + (-4y - 12) = 0
y^2(y + 3) - 4(y + 3) = 0
(y^2 - 4)(y + 3) = 0
(y - 2)(y + 2)(y + 3) = 0
y = 2, -2, -3

Answer 10

mathsmart is absolutely correct. The answers are -2, 2, and -3