Factoring Quadratics
(Leading Coefficient=1)
Multiply \((x+3)(x+4)\)
Multiply \((x+\cancel{3}b)(x+\cancel{4}c)\)
\((x+b)(x+c)\)
=\(x(x+c)+b(x+c)\)
=\(x^2+xc+xb+bc\)
=\(x^2+xb+xc+bc\)
=\(x^2+x(b+c)+bc\)
Thus, \((x+b)(x+c)\)
If \(b+c=7\) and \(bc=6\), then the expression becomes \(x^2+7x+6\).
A common question we must answer in Algebra is if we start with a polynomial, what binomials, if any, would multiply to give us the polynomial.
By Anurag Katyal