Mister Exam
Factor polynomial x^2+x-20
The solution
The perfect square
Let's highlight the perfect square of the square three-member
$$\left(x^{2} + x\right) - 20$$
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
$$a = 1$$
$$b = 1$$
$$c = -20$$
Then
$$m = \frac{1}{2}$$
$$n = - \frac{81}{4}$$
So,
$$\left(x + \frac{1}{2}\right)^{2} - \frac{81}{4}$$
$$\left(x^{2} + x\right) - 20$$
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
$$a = 1$$
$$b = 1$$
$$c = -20$$
Then
$$m = \frac{1}{2}$$
$$n = - \frac{81}{4}$$
So,
$$\left(x + \frac{1}{2}\right)^{2} - \frac{81}{4}$$