1.4.1. Solving quadratics by factorization
\(x+a)(x+b) = x^2+(a+b)x+ab\)
\(c(x-d)(x-e) =0 \implies x=d \text{ or } x=e\)
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You should be able to
Optionally, also see if you can apply these skills to an economic application or look at the additional resources and advanced quizzes at the bottom of the page,
\(x+a)(x+b) = x^2+(a+b)x+ab\)
\(c(x-d)(x-e) =0 \implies x=d \text{ or } x=e\)
Factor Theorem: If \(f(x)\) is a polynomial and \(f(c)=0\), then \((x-c)\) is a factor of \(f(x)\).
Difference of two squares:
\(a^2-b^2 = (a-b)(a+b)\)
You can find practice quizzes on each topic in the links above the tutorial videos, or you can take the diagnostic quiz as many times as you like to. The additional resources gives links to external sources of information and some of these may also have further exercises to try.
Further links and resources 1