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Factorise 2x^2+4x+1 |
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Answer» I see most of the answers here somehow ‘magically’ do a substitution/grouping the elements together without a CLEAR explanation why he/she does so. It seems more or less guessing without a definite way to factorise this. I will show you a small trick I learned several years ago. If you have a calculator, you can easily find the roots of the EQUATION x4+4x3+2x2−4x+1=0 . They are (In this case I use but most advanced calculators have the ability find the roots of a given equation) You may OBSERVE that -2.4142 + 0.41421 = -2 and -2.4142 × 0.41421 = -1 (I round up because the roots are approximate). Recall Vieta's formulas, these two roots are also the roots of the quadratic equation x2+2x−1=0 . Therefore, the EXPRESSION x4+4x3+2x2−4x+1 can be EXPRESSED as a product of x2+2x−1 . By grouping the elements appropriately, you will get: x4+4x3+2x2−4x+1 =(x4+2x3−x2)+(2x3+4x2−2x)+(−x2−2x+1) =x2(x2+2x−1)+2x(x2+2x−1)−(x2+2x−1) =(x2+2x−1)(x2+2x−1) =(x2+2x−1)2 Bear in mind, this technique only works if the 4th degree polynomial is a product of two 2nd degree polynomials and one of them must have real roots. |
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