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Express $2x^5+13x^4+50x^3+82x^2+56x+13$ as a product of five linear factors. The roots of the polynomial may be real or complex.

I had to employ the technique of synthetic division iteratively. I'd like to know if I'm doing this correctly. \begin{align} 2x^5+13x^4+50x^3+82x^2+56x+13 &= (x+1)(2x^4+11x^3+39x^2+43x+13) \\ &= (x+1)^2(3x^3+9x^2+30x+13) \\ &=(x+1)^2\left(x+ \frac 12\right)(2x^2+8x+26) \\ &=(x+1)^2(2x+1)(x^2+4x+13) \\ &=(x+1)^2\left(2x+1\right)(x+(2+3i))(x+(2-3i)) \end{align}

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If you just want to check your answer, you can just use WolframAlpha, which confirms that your factorization is correct. –  JimmyK4542 Jul 1 at 5:24
    
In this sort of problem, they make it feasible by making sure that a fair number of roots are obtainable by using the Rational Roots Theorem. That, or inspection, is presumably what you used. –  André Nicolas Jul 1 at 5:27
    
Yes, I used inspection because it's easier and quicker than cranking out the $\pm \frac pq$ algorithm of the Rational Roots Theorem. –  le gâteau au fromage Jul 1 at 5:43
    
@glace In the last line, the factor $(x-(2+3i))$ must be $(x+(2-3i))$. As it is known that the roots are complex conjugate. (The link given by JimmyK4542 confirms this.) –  Fermat Jul 1 at 6:14
    
If you don't have access to computerised algebra systems, one quick & dirty way of verifying your factorisation is to use an ordinary scientific calculator, plug in a transcendental value for $x$ (like $\pi$) and verify that the LHS is "very close" to the RHS with the discrepancy being fully accountable for by internal rounding errors. This method works because the LHS - RHS forms a polynomial equation with algebraic coefficients, and those only have algebraic roots. For a transcendental number to satisfy it, LHS minus RHS has to be identically zero. Just another tip to make your life easier. –  Deepak Jul 1 at 6:43

1 Answer 1

I like using Newton's method for finding roots in situations like these, so I made an Excel spread sheet.

  • cell A1 contains my initial estimate of the root ($x_k$).
  • cell B1 contains $f$(A1)
  • cell C1 contains $f'$(A1)
  • cell D1 contains my new estimate of the root ($x_{k+1}$), the formula is A1-B1/C1
  • cell D1 is then copied to cell A2

I can enter some random numbers in cell A1 to generate other roots, and the spread sheet will recalculate automatically. By using the values 0 and -2, I found the roots -0.5 and -1.

Unfortunately, this method doesn't identify multiplicities, but that's okay, because I can now reduce $f$ to a cubic by dividing by $(x+1)(x+\frac{1}{2})$ to find $g$ (e.g. $f(x)=(x+1)(x+\frac{1}{2})g(x)$). Once I've found $g'$, I can change columns B and C to reflext the simpler function to find my final Real (and hopefully Rational) factor. I can then use the quadratic formula on whatever is left over.

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