Factoring $3x^2 - 10x + 5$

How can $3x^2 - 10x + 5$ be factored? FOIL seemingly doesn't work (15 has no factors that sum to -10).

• If you take a degree two polynomial at random, what you call FOIL will never work. What works 100% of the time is the discriminant method: en.wikipedia.org/wiki/Quadratic_equation Feb 8, 2013 at 1:30
• factored over which field? Feb 8, 2013 at 1:30
• @IttayWeiss not following you, this is basic pre-calc, nothing complicated. Feb 8, 2013 at 1:32
• factor over the rationals or the reals or perhaps the complex numbers (just ignoring all other possible fields). Generally, when you say "factorize a polynomial" it needs to be clear over which field, as the answer is usually strongly dependent on that. Feb 8, 2013 at 1:35
• @IttayWeiss over the reals in this case Feb 8, 2013 at 1:39

You can always use the general formula for a quadratic equation, if we have $ax^2+bx+c$ then let $$x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}$$ and $$x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$$ then we can write $$ax^2+bx+c=a(x-x_1)(x-x_2)$$ so in particular if we have $3x^2-10x+5$ then $$x_1=\frac{10+\sqrt{40}}{6}$$ and $$x_2=\frac{10-\sqrt{40}}{6}$$ which implies that $$3x^2-10x+5=3(x-\frac{10+\sqrt{40}}{6})(x-\frac{10-\sqrt{40}}{6}).$$

• Good answer, I appreciate the effort. Feb 8, 2013 at 1:53
• @mirai you are welcome. Feb 8, 2013 at 1:54

If you apply the quadratic formula, you get $x=\frac {10 \pm \sqrt{100-60}}6=\frac 16(10 \pm \sqrt{40})$. The $\sqrt {40}$ shows you won't factor it over the rationals. This gives $3(x-\frac 16(10 + \sqrt{40}))(x-\frac 16(10 - \sqrt{40}))$

• Sure, but I'd still need to know the factors of the polynomial. Feb 8, 2013 at 1:32
• @mirai: I have added it Feb 8, 2013 at 1:34

Do the $b^2 - 4ac$ test; in your case $b^2 - 4ac = 100 - 4*3*5 = 40$. This is not a perfect square. You will need to complete the square to factor this.

• I've always wanted to find out, that seems to be an opportunity: are there really people who complete the square rather than using the quadratic formula when the discrimant is nonnegative? Just curious. Feb 8, 2013 at 1:35
• If factoring is annoying, I carry out the test. If I don't get a perfect square, then part of the quadratic formula is computed anyway. Feb 8, 2013 at 1:36
• Ok, thank you for the answer. Feb 8, 2013 at 1:39

$3x^2 - 10x + 5$

Testing the polynomial using the discriminant (the $\;\sqrt{b^2 - 4ac}\;$ portion of the quadratic formula,),

with $$b = -10, \,a=3,\,c = 5$$ gives you $$\sqrt{100 - 4\cdot 3 \cdot 5} = \sqrt{100 - 60} = \sqrt{40} = \sqrt{4\cdot 10} = \sqrt{4}\cdot \sqrt{10} = 2\sqrt{10},$$ which tells you that there aren't any rational, let alone integer, roots.

So we need to find the roots using the entire quadratic formula: For $ax^2 + bx + c$, we get two roots: $$x_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a},\;\; x_2 =\frac{-b - \sqrt{b^2 - 4ac}}{2a}$$ and we already know that $\sqrt{b^2 - 4ac} = 2\sqrt{10}$ for our polynomial:

$$x=\dfrac {10 \pm 2\sqrt{10}}{6} =\dfrac 16 (2\cdot 5 \pm 2\sqrt{10})= 2\cdot\frac 16(5 \pm \sqrt{10}) = \frac13 (5 \pm \sqrt{10})$$

In general, $ax^2+bx+c=a(x-x_1)(x-x_2)$:

So, to express your polynomial in factored form you need $$3x^2 - 10x + 5\quad = \quad 3\left(x-\frac 13(5 + \sqrt{10})\right)\left(x-\frac 13(5 - \sqrt{10})\right)$$

You can try to Conplete the Square in this kind of problem. What you would want to do, is to manipulate $3x^2-10x+5$ into something like this $a^2 \pm 2ab + b^2 = (a \pm b)^2 \quad \mathbf{(1)}$.

First, factor the 3 out, like this $3x^2-10x+5 = 3 \left( x^2 - \dfrac{10}{3}x + \dfrac{5}{3} \right)$, now, look closely inside the brackets, we'll now complete the square.

$x^2$ will stand for $a^2$ in (1), so basically, we'll have $\color{red}{a = x}$ (remember this); now move onto the second term, we'll find $b$, such that, $\dfrac{10}{3}x$ must stand for $2ab$. Since $a = x$, we will have:

\begin{align}2xb &= \dfrac{10}{3}x \\ \Leftrightarrow 2b &= \dfrac{10}{3} \quad \mbox{cancel } x \mbox{ from both sides} \\ \Leftrightarrow b &= \dfrac{5}{3}\end{align}

So, back to the original expression: The red parts, blue parts, and orange parts, although their forms do change, their values remain the same as we move down.

\begin{align} 3x^2-10x+5 &= 3 \left( \color{red}{x^2} - \color{blue}{\dfrac{10}{3}x} + \color{orange}{\dfrac{5}{3}} \right)\\ &= 3 \left( \color{red}{x^2} - \color{blue}{2.x.\dfrac{5}{3}} + \color{orange}{\dfrac{5}{3}} \right) \quad \mbox{make the }a^2 + 2ab \mbox{ part appear}\\ &= 3 \left( \color{red}{x^2} - \color{blue}{2.x.\dfrac{5}{3}} + \color{orange}{\left( \dfrac{5}{3} \right)^2 - \left( \dfrac{5}{3} \right)^2 + \dfrac{5}{3}} \right) \quad \begin{array}{l} \mbox{make the }b^2 \mbox{ part} \\ \mbox{appear, and in other to keep}\\\mbox{the orange part the same,}\\ \mbox{we'll add and subtract simultaneously}\end{array}\\ &= 3 \left[ \left( \color{red}{x^2} - \color{blue}{2.x.\dfrac{5}{3}} + \color{orange}{\left( \dfrac{5}{3} \right)^2} \right) \color{orange}{- \left( \dfrac{5}{3} \right)^2 + \dfrac{5}{3}} \right] \quad \mbox{re-grouping}\\ &= 3 \left[ \left( \color{red}{x^2} - \color{blue}{2.x.\dfrac{5}{3}} + \color{orange}{\left( \dfrac{5}{3} \right)^2} \right) \color{orange}{- \dfrac{25}{9} + \dfrac{5}{3}} \right]\\ &= 3 \left[ \left( \color{red}{x^2} - \color{blue}{2.x.\dfrac{5}{3}} + \color{orange}{\left( \dfrac{5}{3} \right)^2} \right) \color{orange}{- \dfrac{10}{9}} \right]\\ &= 3 \left[ \left( x - \dfrac{5}{3} \right)^2 \color{orange}{- \dfrac{10}{9}} \right] \quad \mbox{apply } \mathbf{(1)}\\ &= 3 \left[ \left( x - \dfrac{5}{3} \right)^2 \color{orange}{- \left(\dfrac{\sqrt{10}}{3} \right)^2} \right]\\ &= 3 \left[ \left( x - \dfrac{5}{3} - \dfrac{\sqrt{10}}{3} \right) \left( x - \dfrac{5}{3} + \dfrac{\sqrt{10}}{3} \right) \right] \quad \mbox{applying difference of squares formula}\\ &= 3 \left( x - \dfrac{5+\sqrt{10}}{3}\right) \left( x + \dfrac{\sqrt{10} - 5}{3}\right)\quad \mbox{simplify it a bit} \end{align}