Find the roots of y³+6y²+11y+6=0 and y³+6y²+12y+8=0

[danny12345]

1)y^3+6y^2+11y+6=0
2)y^3+6y^2+12y+8=0
find it's root and tell me how you obtained it.

 

[Guest2]

Use the procedure outlined on this page.

 

[MarkFL]

1)y^3+6y^2+11y+6=0
2)y^3+6y^2+12y+8=0
find it's root and tell me how you obtained it.

I'll work the first one, and then see if you can use the same procedure to work the second one.

1.) Let:

$$P(y)=y^3+6y^2+11y+6$$

Now, the rational roots theorem tells us that if $P$ has any rational roots, they must be come from a list of numbers generated by taking all of the factors of the constant term and dividing them by all of the factors of the leading coefficient. Since the leading coefficient is 1, we need only look at the factors of 6. Since all of the terms in $P$ have a " sign in front of them, we know any rational roots must be negative so that the terms with odd exponents will be negative and will cancel with the positive terms with even exponents. So, we look at this list as possible candidates:

$$y\in\{-1,-2,-3,-6\}$$

Proceeding systematically, we find:

$$P(-1)=(-1)^3+6(-1)^2+11(-1)+6=-1+6-11+6=0$$ and so we know -1 is a root.

$$P(-2)=(-2)^3+6(-2)^2+11(-2)+6=-8+24-22+6=0$$ and so we know -2 is a root.

$$P(-3)=(-3)^3+6(-3)^2+11(-3)+6=-27+54-33+6=0$$ and so we know -3 is a root.

Because a cubic polynomial can have no more than 3 roots, we are done. We can now state:

$$P(y)=y^3+6y^2+11y+6=(y+1)(y+2)(y+3)$$

If we had only found 1 rational root, then we could use polynomial division to obtain the other roots from the resulting quadratic quotient (using the quadratic formula or completing the square).

 

[Wilmer]

y^3+6y^2+11y+6=0
y^3+6y^2+12y+8=0

Well, if I got that on a timed test,
I'd simply subtract the equations
to get y+2 = 0, so y = -2

Would you give me a pass mark, Mark :)

 

[MarkFL]

y^3+6y^2+11y+6=0
y^3+6y^2+12y+8=0

Well, if I got that on a timed test,
I'd simply subtract the equations
to get y+2 = 0, so y = -2

Would you give me a pass mark, Mark :)

I don't think the two equations are simultaneous (even though the OP said "find its root" as if the two equations were one entity)...so if my interpretation is correct and unless you used this in lieu of the rational roots theorem to find a root as a starting point for both, I would have to deduct some points for not finding ALL roots of both equations. :)

 

[Wilmer]

I don't think the two equations are simultaneous (even though the OP said "find its root" as if the two equations were one entity)...so if my interpretation is correct and unless you used this in lieu of the rational roots theorem to find a root as a starting point for both, I would have to deduct some points for not finding ALL roots of both equations. :)

YA! On a timed test, that's all I'd want.