#rough https://www.mashupmath.com/blog/how-to-factor-polynomials To factor a [[polynomial]] means to rewrite the expression as a product of factors. Recognizing patterns is an important skill in factoring a polynomial. ## ac test of factorability If in polynomials on the form $\begin{align} ax^2+bx+c && \text{or} && ax^2 +bxy + cy^2 \end{align}$ the product $ac$ has two integer factors $p$ and $q$ whose sum is the coefficient $b$ of the middle term; that is, if integers $p$ and $q$ exist such that $\begin{align} pq = ac && \text{and} && p + q = b \end{align}$ then the polynomials have first-degree factors with integer coefficients. If not, we can say that the polynomial is **not factorable in the integers**. Create a factor table for the product $ac$ to enumerate all factors, and test each one to see if the sum is $b$. For example, in the polynomial $3x^2 + 8xy + 4y^2$, the product $ac = 12$ which has factors | p | q | sum | | --- | --- | --- | | 1 | 12 | 13 | | 2 | 6 | 8 | | 3 | 4 | 7 | The polynomial is factorable. We can split the middle term using $b = p + q$ and complete the factoring by grouping. $\begin{align} 3x^2 + 8xy + 4y^2 &= 3x^2 + 2xy + 6xy + 4y^2 \\ &= (3x^2 + 2xy) + (6xy + 4y^2) \\ &= x(3x + 2y) + 2y(3x + 2y) \\ &= (x + 2y)(3x + 2y) \end{align}$ ## special factoring formulas Some common factors include - **Perfect square:** $u^2 + 2uv + v^2 = (u + v)^2$ - **Perfect square (difference):** $u^2 - 2uv + v^2 = (u - v)^2$ - **Difference of squares:** $u^2 - v^2 = (u-v)(u+v)$ - **Difference of cubes:** $u^3 - v^3 = (u - v)(u2 + uv + v^2)$ - **Sum of cubes:** $u^3 + v^3 = (u + v)(u^2 - uv + v^2)$ ## factoring strategies ### group like terms Group (add or subtract) any terms of the same degree (those raised to the same power). ### greatest common factor If all terms share a common factor, extract it. ### difference of two squares The difference of two squares can be factored as $(a^2 - b^2) = (a+b)(a-b)$ ## factoring a trinomial where a=1 For a polynomial with three terms where the coefficient of the first term is $1$, if the coefficient for $x$ is the sum of two numbers whose product is the constant, you can factor the polynomial. For example $(x^2 + 6x + 8) = (x + 2)(x + 4)$ ## Factoring a squared polynomial When factoring a polynomial that is squared, square both factors. $f(x) = (x^2 - 1)^2 = (x + 1)^2(x - 1)^2$