Factorize abx2−(a2+b2)x+ab

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ron4d3ozip7 · Accepted Answer

Step 1Using the Quadratic Formula is more general but here is an alternative method:We haveabx2−(a2+b2)x+ab=0⇔x2−a2+b2abx+1=0⇔x2−(ab+ba)x+1=0Let r1 and r2 be the (possibly equal, possibly complex) roots of the quadratic function x2−(ab+ba)x+1Then their product is 1 and their sum is ab+ba. This showns that the roots are ba and abTherefore, x2−(ab+ba)x+1=(x−ab)(x−ba) soabx2−(a2+b2)x+ab=ab(x−ab)(x−ba)=(ax−b)(bx−a)

Ariel Cantrell · Accepted Answer

Step 1The polynomial Cx2+Bx+A is referred to as the reciprocal polynomial to Ax2+Bx+C, as its zeroes are the reciprocals of the latter polynomial. In the same vein, a polynomial Ax2+Bx+A is called palindromic and has the property that if r is a zero, then 1r is as well. (This applies to polynomials of higher degree as well as quadratics.)This permits us to writeab×x2−(a2+b2)×x+ab=ab×[x2−a2+b2ab×x+1]=ab×[x2−(ba+ab)×x+1]=ab×(x−r)×(x−1r)=ab×[x2−(r+1r)×x+1]The reciprocal nature of the two zeroes is thus confirmed and the factorization proceeds as in Taladris&#039;s answer.

Factorize \(\displaystyle{a}{b}{x}^{{2}}-{\left({a}^{{2}}+{b}^{{2}}\right)}{x}+{a}{b}\)

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