Factorising

[ApolloAstro]

I have been presented with the question

f(x) = 2x³ + 3x²- 12x - 15

and I have been asked to find the stationary points of f, inculding their y-coordinates

so far I have got:

f(x) = 2x³ + 3x²- 12x - 15
f'(x) = 6x²+ 6x - 12

the common multiple I have found is 2;

2(3x + 3x - 6)

I need to factorise it, but cannot find any soultions, can anyone help please.

 

[srmichael]

I have been presented with the question

f(x) = 2x³ + 3x²- 12x - 15

and I have been asked to find the stationary points of f, inculding their y-coordinates

so far I have got:

f(x) = 2x³ + 3x²- 12x - 15
f'(x) = 6x²+ 6x - 12

the common multiple I have found is 2;

2(3x + 3x - 6)

I need to factorise it, but cannot find any soultions, can anyone help please.

First, your first term inside the parenthesis has a typo. It should be 3x². Second, try factoring out a 6 instead of a 2. It may be more apparent to you how to factor.

 

[HallsofIvy]

I have been presented with the question

f(x) = 2x³ + 3x²- 12x - 15

and I have been asked to find the stationary points of f, inculding their y-coordinates

so far I have got:

f(x) = 2x³ + 3x²- 12x - 15
f'(x) = 6x²+ 6x - 12

the common multiple I have found is 2;

2(3x + 3x - 6)

And now all coefficients, 3, 3, and -6, are divisible by 3. In fact, I am surprised that you did not see that 6 is a factor to begin with:
\(\displaystyle f'(x)= 6(x^2+ x- 2)\)

No, the only possible rational roots of \(\displaystyle x^2+ x- 2= 0\) are the factors of -2: 1, -1, 2, and -2. Try those and you should be able to determine two roots immediately. l

But even if there were no integer or rational roots, if you are taking Calculus, you should already have learned to "complete the square" and use the "quadratic formula".

I need to factorise it, but cannot find any soultions, can anyone help please.