Mister Exam
Factor polynomial 2*x^2-x-1
The solution
The perfect square
Let's highlight the perfect square of the square three-member
$$\left(2 x^{2} - x\right) - 1$$
To do this, let's use the formula
$$a x^{2} + b x + c = a \left(m + x\right)^{2} + n$$
where
$$m = \frac{b}{2 a}$$
$$n = \frac{4 a c - b^{2}}{4 a}$$
In this case
$$a = 2$$
$$b = -1$$
$$c = -1$$
Then
$$m = - \frac{1}{4}$$
$$n = - \frac{9}{8}$$
So,
$$2 \left(x - \frac{1}{4}\right)^{2} - \frac{9}{8}$$
$$\left(2 x^{2} - x\right) - 1$$
To do this, let's use the formula
$$a x^{2} + b x + c = a \left(m + x\right)^{2} + n$$
where
$$m = \frac{b}{2 a}$$
$$n = \frac{4 a c - b^{2}}{4 a}$$
In this case
$$a = 2$$
$$b = -1$$
$$c = -1$$
Then
$$m = - \frac{1}{4}$$
$$n = - \frac{9}{8}$$
So,
$$2 \left(x - \frac{1}{4}\right)^{2} - \frac{9}{8}$$
Factorization
[src]
(x + 1/2)*(x - 1)
$$\left(x - 1\right) \left(x + \frac{1}{2}\right)$$
(x + 1/2)*(x - 1)
Combinatorics
[src]
(1 + 2*x)*(-1 + x)
$$\left(x - 1\right) \left(2 x + 1\right)$$
(1 + 2*x)*(-1 + x)
Combining rational expressions
[src]
-1 + x*(-1 + 2*x)
$$x \left(2 x - 1\right) - 1$$
-1 + x*(-1 + 2*x)