Having a hard time locating answers. Eventually I get them, but it's painstaking.
Take this question
12x2 + 11x + 2 = 0
So after a tonne of inserting different fractions into the equation (like over a whole page) I found the answers to be - x is either (-1/4) or (-2/3)
So, is there any way at all to find these answers quicker, without having to just blindly throw in fraction after fraction after fraction to find what works?
Thanks for the help
There certainly is! This is a polynomial, specifically a quadratic (second degree). We actually have a formula, the quadratic formula, that tells us the exact solutions for a quadratic equation. I'll show you the formula a little later, for now we'll look at another way of solving it (the quadratic formula is overkill for this particular example).
For "nice" polynomials like the one you provided, we can often "factor" them using the basic rules of algebra. This lets us find their solutions without having to go through more strenuous work. For your quadratic, we can factor as follows:
\(\displaystyle \begin{align*}
12x^2 + 11x + 2 &= 12x^2 + 8x + 3x + 2 \\
&= (12x^2 + 8x) + (3x + 2) \\
&= 4x(3x + 2) + (3x + 2) \\
&= (4x + 1)(3x + 2).
\end{align*}\)
The first step was to split up the \(\displaystyle 11x\) as \(\displaystyle 11x = 8x + 3x\), then I factored by grouping, pulling the \(\displaystyle 4x\) out of the first group using the distributive law in reverse. The last step was to factor out the common factor of \(\displaystyle 3x+2\). How did I know to split the \(\displaystyle 11x\) into \(\displaystyle 8x\) and \(\displaystyle 3x\)? Well, it's a standard factoring technique for trinomials (polynomials with three terms): you take the leading coefficient 12 and multiply by the constant term 2 to get 24, and then you look for two numbers whose sum is the middle coefficient 11 and whose product is 24. 8 and 3 work.
From here, we get that \(\displaystyle (4x+1)(3x+2) = 0\). And when multiple numbers have a product of zero, that means one of the numbers must be zero. So either \(\displaystyle 4x+1=0\) or \(\displaystyle 3x+2=0\). In the first case we get \(\displaystyle x = -1/4\) and in the second \(\displaystyle x = -2/3\).
Now, earlier I mentioned the quadratic formula. This can always be used to solve a quadratic, and it'll even find the complex solutions that aren't real numbers. The formula for the solution of \(\displaystyle ax^2 + bx + c = 0\) is
\(\displaystyle \displaystyle
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
\)
So in your example, \(\displaystyle a = 12, b = 11,\) and \(\displaystyle c = 2\). So you can just plug those numbers in and get an answer. What about higher degree polynomials? It turns out that there is a (really long and complicated) formula for solving a cubic equation \(\displaystyle ax^3 + bx^2 + cx + d = 0\), but it's so long and unwieldy that you're better off trying to either factor or use a computer algebra system to do the tedious work. There's a formula for a 4th-degree polynomial too, and it's even longer and more complicated than the cubic.
We don't know of a 5th-degree formula, and you might think that there is one out there waiting to be discovered. Remarkably, it has been proven that the solutions to the general 5th-degree polynomial equation and all higher degrees cannot be expressed by a general formula in radicals in the same way that the quadratic and cubic and quartic can! Isn't that interesting? We can still solve particular equations, and there are factoring techniques that can help us find roots of higher-degree polynomials, but there is no general formula that works for every case above \(\displaystyle n=4\).
Definitely read up on factoring, especially if you're working with rational expressions. The rational root theorem can also be helpful.
