Mister Exam
Other calculators
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How to use it?
- Factor squared:
- -y^4+y^2+2
- y^4-y^2-13
- -y^4-8*y^2-2
- -y^4-8*y^2-4
- Factor polynomial:
- z^3+i^3
- z^3+8*i
- z^3-6*z^2+5*z+12
- z^3+5*z^2+4*z-10
- Least common denominator:
- (z^2-2*t+2*t1+exp(t1/t)*(2*t-z^2))*(z^2-2*t+2*t2+exp(t2/t)*(2*t-z^2))
- ((z^2+1)^2/(4*z^2))/(13+12*((z^2+1)/2*z*z))
- -z/1-z/2
- y*(y+((1/(x+(x^2+y^2)^(1/2))*(1+(2*x/(2*(x^2+y^2)^(1/2)))))))-(y/(x^2+y^2)^(1/2))
- How do you in partial fractions?:
- (1-x/(1+x))/(1+x)^3
- (x^2+6*x+8)/(x+4)
- (x^2-4)/(x-2)^2
- -3/(x*x^3)
- Graphing y =:
- x^2+x-1
- Integral of d{x}:
-
x^2+x-1
- Canonical form:
- x^2+x-1
-
Identical expressions
- x^ two +x- one
- x squared plus x minus 1
- x to the power of two plus x minus one
- x2+x-1
- x²+x-1
- x to the power of 2+x-1
-
Similar expressions
- x^2+x+1
- x^2-x-1
Factor x^2+x-1 squared
The solution
The perfect square
Let's highlight the perfect square of the square three-member
$$\left(x^{2} + x\right) - 1$$
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
$$a = 1$$
$$b = 1$$
$$c = -1$$
Then
$$m = \frac{1}{2}$$
$$n = - \frac{5}{4}$$
So,
$$\left(x + \frac{1}{2}\right)^{2} - \frac{5}{4}$$
$$\left(x^{2} + x\right) - 1$$
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
$$a = 1$$
$$b = 1$$
$$c = -1$$
Then
$$m = \frac{1}{2}$$
$$n = - \frac{5}{4}$$
So,
$$\left(x + \frac{1}{2}\right)^{2} - \frac{5}{4}$$
Factorization
[src]
/ ___\ / ___\ | 1 \/ 5 | | 1 \/ 5 | |x + - - -----|*|x + - + -----| \ 2 2 / \ 2 2 /
$$\left(x + \left(\frac{1}{2} - \frac{\sqrt{5}}{2}\right)\right) \left(x + \left(\frac{1}{2} + \frac{\sqrt{5}}{2}\right)\right)$$
(x + 1/2 - sqrt(5)/2)*(x + 1/2 + sqrt(5)/2)