\pi \cdot \pi

Generalization

Factoring Quadratics

(Leading Coefficient=1)

 

Multiply \((x+3)(x+4)\)

\pi \cdot \pi

Multiply \((x+\cancel{3}b)(x+\cancel{4}c)\)

\pi \cdot \pi

\((x+b)(x+c)\)

Multiply \((x+\cancel{3}b)(x+\cancel{4}c)\)

\pi \cdot \pi

\((x+b)(x+c)\)

=\(x(x+c)+b(x+c)\)

Multiply \((x+\cancel{3}b)(x+\cancel{4}c)\)

\pi \cdot \pi

\((x+b)(x+c)\)

=\(x(x+c)+b(x+c)\)

=\(x^2+xc+xb+bc\)

Multiply \((x+\cancel{3}b)(x+\cancel{4}c)\)

\pi \cdot \pi

\((x+b)(x+c)\)

=\(x(x+c)+b(x+c)\)

=\(x^2+xc+xb+bc\)

Multiply \((x+\cancel{3}b)(x+\cancel{4}c)\)

=\(x^2+xb+xc+bc\)

\pi \cdot \pi

\((x+b)(x+c)\)

=\(x(x+c)+b(x+c)\)

=\(x^2+xc+xb+bc\)

Multiply \((x+\cancel{3}b)(x+\cancel{4}c)\)

=\(x^2+xb+xc+bc\)

=\(x^2+x(b+c)+bc\)

\pi \cdot \pi

Thus, \((x+b)(x+c)\)

=\(x^2+x(b+c)+bc\)

\pi \cdot \pi

Multiply \((x+\cancel{3}b)(x+\cancel{4}c)\)

Thus, \((x+b)(x+c)\)

=\(x^2+x(b+c)+bc\)

\pi \cdot \pi

If \(b+c=7\) and \(bc=6\), then the expression becomes \(x^2+7x+6\).

Multiply \((x+\cancel{3}b)(x+\cancel{4}c)\)

\pi \cdot \pi

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.

Copy of Factoring Quadratics with a Leading Coefficient of 1 Generalization

By Anurag Katyal

Copy of Factoring Quadratics with a Leading Coefficient of 1 Generalization

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