I have $3 = -x^2 + 4x$ and I need to solve it by factoring. According to wolframalpha the solution is $x_1 = 1, x_2 = 3$. \begin{align*} 3 & = -x^2 + 4x\\ x^2-4x+3 & = 0 \end{align*}
According to wolframalpha $(x-3) (x-1) = 0$ is the equation factored, which allows me to solve it, but how do I get to this step?
On
Move everything to one side then factor. Factoring something like this is just trial and error. You need to find two numbers whose product is $3$ and whose sum is $-4$. So poking around a bit you should come up with $-3$ and $-1$. So it factors as: $x^2-4x+3=(x-3)(x-1)$
On
Here's a similar problem. Let's solve $$x^2 + 6 = 7x$$ by factoring.
First, we need to write this in standard form: $$x^2 - 7x + 6 = 0$$
Then we seek two numbers whose sum is $-7$, and whose product is $6$. The two such numbers are $-1$ and $-6$.
At this point, you could just jump straight to $(x - 1)(x-6) = 0$, but if you have never seen this before, it might help to rewrite the middle term and factor by grouping. To wit:
$$x^2 - 7x + 6 = x^2 -1x - 6x + 6$$
We then write this as $$x(x-1) -6(x-1) = (x-6)(x-1)$$
From here, we set each factor equal to zero.
On
Solving by factoring means reducing the polynomial to a lower degree by dividing it by factors $(x-x_i)$ for already known roots $x_i$.
The reason for this is that the lower degree polynomial might be easier to solve. Here this is not necessary, as a order $2$ polynomial is already solvable with moderate effort.
How to know roots? For this kind of task one should always try the $0, \pm 1, \pm 2$, which here gives $x_1 = 1$.
What is left is the division: \begin{align} & x^2 - 4x + 3 : x - 1 = x - 3 \\ -&(x^2-x) \\ = & -3x+3 = -3(x-1) \end{align} So we have the factored polynomial $(x+1)(x-3)$.
On
$$ 3 = -x^2 +4x \implies \ x^2 -4x+3 = 0$$
$$ x^2 - 3x - x +3 = 0$$
$$ x(x-3)-1(x-3)$$ $$ (x-3)(x-1) = 0$$ $$x = 1, x = 3$$
For more on the methods, visit this link.
Alternatively, you could use the quadratic formula:
$$x = \frac{-b \pm\sqrt{b^2 - 4ac}}{2a}$$
This was obtained from the quadratic equation:
$$ax^2 + bx +c = 0$$ where $a = 1, b = -4, c = 3$
You moved everything onto the same side of the equation, which is a great start!
The next step is to understand factorization of quadratic polynomials.
Suppose you have a polynomial of the form
$$x^2 + cx + d \tag{1} $$
You want something of the form
$$ (x+a)(x+b) $$
When you expand out $(x+a) (x+b)$, you get $$ (x+a)(x+b) = x^2 + (a+b)x + ab \tag{2}$$
Now, when we compare the coefficients of (2) to (1), we see that
$$ a+b = c $$ $$ ab = d $$
That is, we need to find two numbers $a$ and $b$ that add up to $c$ and multiply into $d$.
In our case, $c$ is -4 and $d$ is 3. Now we have to think about it for a bit and do a bit of guessing and checking, but you should be able to see that $a=-3$ and $b =-1$ meets this criteria.
-3+(-1) = -4, and (-3)*(-1)=3.