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

AI Thread Summary
Factoring the polynomial \(6x^4+17x^3-24x^2-53x+30\) can be achieved without the rational root theorem by grouping terms and using trial and error. The polynomial can be rewritten as \((x+2)(x+3)(2x-1)(3x-5)\) through strategic grouping and simplification. The process often involves guesswork to identify pairs of numbers that facilitate factoring. One participant mentioned using Wolfram Alpha for assistance in determining the factors. Overall, the discussion highlights the combination of analytical methods and computational tools in polynomial factorization.
paulmdrdo1
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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
 
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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:

$$6x^4+17x^3-24x^2-53x+30=(x+2)(x+3)(2x-1)(3x-5)$$
 
MarkFL said:
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?
 
paulmdrdo said:
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
 
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