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

Guadalupe Glass

Guadalupe Glass

Answered question

2022-04-02

Factorize abx2(a2+b2)x+ab

Answer & Explanation

ron4d3ozip7

ron4d3ozip7

Beginner2022-04-03Added 9 answers

Step 1
Using the Quadratic Formula is more general but here is an alternative method:
We have
abx2(a2+b2)x+ab=0x2a2+b2abx+1
=0x2(ab+ba)x+1=0
Let r1 and r2 be the (possibly equal, possibly complex) roots of the quadratic function x2(ab+ba)x+1
Then their product is 1 and their sum is ab+ba. This showns that the roots are ba and ab
Therefore, x2(ab+ba)x+1=(xab)(xba) so
abx2(a2+b2)x+ab=ab(xab)(xba)
=(axb)(bxa)
Ariel Cantrell

Ariel Cantrell

Beginner2022-04-04Added 8 answers

Step 1
The 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 write
ab×x2(a2+b2)×x+ab
=ab×[x2a2+b2ab×x+1]
=ab×[x2(ba+ab)×x+1]
=ab×(xr)×(x1r)
=ab×[x2(r+1r)×x+1]
The reciprocal nature of the two zeroes is thus confirmed and the factorization proceeds as in Taladris's answer.

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