How can I factor a polynomial without using the rational root theorem?

[paulmdrdo1]

I'm just curious as to how to go about factoring a polynomial like this one $6x^4+17x^3-24x^2-53x+30$ without using rational root theorem?

Thanks

 

[MarkFL]

You could write:

$$6x^4+17x^3-24x^2-53x+30=\left(6x^4+12x^3\right)+\left(5x^3+10x^2\right)-\left(34x^2+68x\right)+\left(15x+30\right)=6x^3(x+2)+5x^2(x+2)-34x(x+2)+15(x+2)=(x+2)\left(6x^3+5x^2-34x+15\right)$$

$$6x^3+5x^2-34x+15=\left(6x^3+18x^2\right)-\left(13x^2+39x\right)+\left(5x+15\right)=6x^2(x+3)-13x(x+3)+5(x+3)=(x+3)\left(6x^2-13x+5\right)$$

$$6x^2-13x+5=(2x-1)(3x-5)$$

And so we have:

$$6x^4+17x^3-24x^2-53x+30=(x+2)(x+3)(2x-1)(3x-5)$$

 

[paulmdrdo1]

You could write:

$$6x^4+17x^3-24x^2-53x+30=\left(6x^4+12x^3\right)+\left(5x^3+10x^2\right)-\left(34x^2+68x\right)+\left(15x+30\right)=6x^3(x+2)+5x^2(x+2)-34x(x+2)+15(x+2)=(x+2)\left(6x^3+5x^2-34x+15\right)$$

$$6x^3+5x^2-34x+15=\left(6x^3+18x^2\right)-\left(13x^2+39x\right)+\left(5x+15\right)=6x^2(x+3)-13x(x+3)+5(x+3)=(x+3)\left(6x^2-13x+5\right)$$

$$6x^2-13x+5=(2x-1)(3x-5)$$

And so we have:

$$6x^4+17x^3-24x^2-53x+30=(x+2)(x+3)(2x-1)(3x-5)$$

What technique did you use to determine the pair of numbers to be used to rewrite the expression? Was it by trial and error?

 

[MarkFL]

What technique did you use to determine the pair of numbers to be used to rewrite the expression? Was it by trial and error?

Yes, there is generally guesswork involved in factoring polynomials. What I actually did was used W|A to factor it, and then constructed my post based on that. :D