I came across this in a friend's 12th grade math homework and couldn't solve it. I want to factor the following trinomial: $$3x^2 -8x + 1.$$
How to solve this is far from immediately clear to me, but it is surely very easy. How is it done?
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this communityI came across this in a friend's 12th grade math homework and couldn't solve it. I want to factor the following trinomial: $$3x^2 -8x + 1.$$
How to solve this is far from immediately clear to me, but it is surely very easy. How is it done?
We have the following theorem (ViƩte formulae) for the quadratic case:
If $\,\alpha\,,\,\beta\,$ are roots (i.e., solutions) of the quadratic equation $\,ax^2+bx+c=0\,$ , then we have that$$ax^2+bx+c=a(x-\alpha)(x-\beta)$$
This is the reason why Eugene hinted you at using the quadratic formula
A standard way of factorizing, when it is hard to guess the factors, is by completing the square. \begin{align} 3x^2 - 8x + 1 & = 3 \left(x^2 - \dfrac83x + \dfrac13 \right)\\ & (\text{Pull out the coefficient of $x^2$})\\ & = 3 \left(x^2 - 2 \cdot \dfrac43 \cdot x + \dfrac13 \right)\\ & (\text{Multiply and divide by $2$ the coefficient of $x$})\\ & = 3 \left(x^2 - 2 \cdot \dfrac43 \cdot x + \left(\dfrac43 \right)^2 - \left(\dfrac43 \right)^2 + \dfrac13 \right)\\ & (\text{Add and subtract the square of half the coefficient of $x$})\\ & = 3 \left(\left(x - \dfrac43 \right)^2 - \left(\dfrac43 \right)^2 + \dfrac13 \right)\\ & (\text{Complete the square})\\ & = 3 \left(\left(x - \dfrac43 \right)^2 - \dfrac{16}9 + \dfrac13 \right)\\ & = 3 \left(\left(x - \dfrac43 \right)^2 - \dfrac{16}9 + \dfrac39 \right)\\ & = 3 \left(\left(x - \dfrac43 \right)^2 - \dfrac{13}9\right)\\ & = 3 \left(\left(x - \dfrac43 \right)^2 - \left(\dfrac{\sqrt{13}}3 \right)^2\right)\\ & = 3 \left(x - \dfrac43 + \dfrac{\sqrt{13}}3\right) \left(x - \dfrac43 - \dfrac{\sqrt{13}}3\right)\\ & (\text{Use $a^2 - b^2 = (a+b)(a-b)$ to factorize}) \end{align} The same idea works in general. \begin{align} ax^2 + bx + c & = a \left( x^2 + \dfrac{b}ax + \dfrac{c}a\right)\\ & = a \left( x^2 + 2 \cdot \dfrac{b}{2a} \cdot x + \dfrac{c}a\right)\\ & = a \left( x^2 + 2 \cdot \dfrac{b}{2a} \cdot x + \left( \dfrac{b}{2a}\right)^2 - \left( \dfrac{b}{2a}\right)^2 + \dfrac{c}a\right)\\ & = a \left( \left( x + \dfrac{b}{2a}\right)^2 - \left( \dfrac{b}{2a}\right)^2 + \dfrac{c}a\right)\\ & = a \left( \left( x + \dfrac{b}{2a}\right)^2 - \dfrac{b^2}{4a^2} + \dfrac{c}a\right)\\ & = a \left( \left( x + \dfrac{b}{2a}\right)^2 - \left(\dfrac{b^2-4ac}{4a^2} \right)\right)\\ & = a \left( \left( x + \dfrac{b}{2a}\right)^2 - \left(\dfrac{\sqrt{b^2-4ac}}{2a} \right)^2\right)\\ & = a \left( x + \dfrac{b}{2a} + \dfrac{\sqrt{b^2-4ac}}{2a}\right) \left( x + \dfrac{b}{2a} - \dfrac{\sqrt{b^2-4ac}}{2a}\right)\\ \end{align}
There's an approach called (by some) the "$ac$ method". Suppose $\alpha,\beta,\gamma,\delta$ are some constants, and consider the expansion $$(\alpha x+\beta)(\gamma x+\delta)=\alpha\gamma x^2+(\alpha\delta+\beta\gamma)x+\beta\delta=ax^2+bx+c.$$ Note that $\alpha\delta$ and $\beta\gamma$ are factors of $ac=\alpha\beta\gamma\delta$ and that $b=\alpha\delta+\beta\gamma$. In fact, they are paired factors of $ac$, in that $\alpha\delta\cdot\beta\gamma=ac$. The idea of the $ac$ method of factoring is to find such paired factors of $ac$ whose sum is $b$.
To see how this can be useful, consider the example $3x^2+13x-10$. Here, $ac=-30$ and $b=13$. We need to find two numbers that add up to $13$ and multiply to get $-30$. The only pair that works is $15$ and $-2$. Thus, we can rewrite $13$ as $15-2$ or $-2+15$. How is this useful? Well, $$3x^2+13x-10=3x^2+15x-2x-10=3x(x+5)-2(x+5)=(3x-2)(x+5)$$ and $$3x^2+13x-10=3x^2-2x+15x-10=x(3x-2)+5(3x-2)=(x+5)(3x-2),$$ so either way, we obtain our factorization without too much difficulty.
In this case, unfortunately, we have $ac=3$ and $b=-8$, so there is no obvious pair of factors! Such a pair does exist, namely $-4+\sqrt{13}$ and $-4-\sqrt{13}$, but even knowing the pair may not make the factorization simple in all cases (in this case it is, because $c=1$).
In general, if trying to factor something intractible to the $ac$ method, I recommend completing the square, as Marvis has demonstrated so well.