Factoring Quadratics
(Leading Coefficient=1)
Multiply (x+3)(x+4)
Multiply (x+3b)(x+4c)
(x+b)(x+c)
Multiply (x+3b)(x+4c)
(x+b)(x+c)
=x(x+c)+b(x+c)
Multiply (x+3b)(x+4c)
(x+b)(x+c)
=x(x+c)+b(x+c)
=x2+xc+xb+bc
Multiply (x+3b)(x+4c)
(x+b)(x+c)
=x(x+c)+b(x+c)
=x2+xc+xb+bc
Multiply (x+3b)(x+4c)
=x2+xb+xc+bc
(x+b)(x+c)
=x(x+c)+b(x+c)
=x2+xc+xb+bc
Multiply (x+3b)(x+4c)
=x2+xb+xc+bc
=x2+x(b+c)+bc
Thus, (x+b)(x+c)
=x2+x(b+c)+bc
Multiply (x+3b)(x+4c)
Thus, (x+b)(x+c)
=x2+x(b+c)+bc
If b+c=7 and bc=6, then the expression becomes x2+7x+6.
Multiply (x+3b)(x+4c)
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.