Mister Exam
Other calculators
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How to use it?
- Factor polynomial:
- x^6*y^9-64*x^3
- y^5+y^3+y^2+1
- x^7+x^5+1
- m^3+m^2+m+1-m^4-m^3-m^2-m-1
- Least common denominator:
- factorial(n)/factorial(n+1)-(n-1)/factorial(n)
- a3*b3/a3-a2*b*b2-a2/4*ab2
- (x-1)*(1/(2*(x-1))-(x+1)/(2*(x-1)^2))/(x+1)
- sin(2*pi+x)+((3*pi)/2-x)+cos(x+pi/2)
- Factor squared:
- -q^4-4*q^2+5
- -9*p^2-5*p-4
- x^2-x*a+4*a^2
- 8*x^4-14*x^2-3
- How do you in partial fractions?:
- (x^2+2x-15)/(x+5)
- (4x^2-3x+1)/(x^4+x^2)
- ((p+3)/(p^2-2p))*((4p-8)/(p+3))
- a^3-a^2-a-2/(a^3+1)*a^3-2*a^2+2*a-1/(a^3+a^2+a)*a^2+a/(a^2-3*a+2)
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Identical expressions
- -q^ four - four *q^ two + five
- minus q to the power of 4 minus 4 multiply by q squared plus 5
- minus q to the power of four minus four multiply by q to the power of two plus five
- -q4-4*q2+5
- -q⁴-4*q²+5
- -q to the power of 4-4*q to the power of 2+5
- -q^4-4q^2+5
- -q4-4q2+5
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Similar expressions
- -q^4+4*q^2+5
- -q^4-4*q^2-5
- q^4-4*q^2+5
Factor -q^4-4*q^2+5 squared
The solution
The perfect square
Let's highlight the perfect square of the square three-member
$$\left(- q^{4} - 4 q^{2}\right) + 5$$
To do this, let's use the formula
$$a q^{4} + b q^{2} + c = a \left(m + q^{2}\right)^{2} + n$$
where
$$m = \frac{b}{2 a}$$
$$n = \frac{4 a c - b^{2}}{4 a}$$
In this case
$$a = -1$$
$$b = -4$$
$$c = 5$$
Then
$$m = 2$$
$$n = 9$$
So,
$$9 - \left(q^{2} + 2\right)^{2}$$
$$\left(- q^{4} - 4 q^{2}\right) + 5$$
To do this, let's use the formula
$$a q^{4} + b q^{2} + c = a \left(m + q^{2}\right)^{2} + n$$
where
$$m = \frac{b}{2 a}$$
$$n = \frac{4 a c - b^{2}}{4 a}$$
In this case
$$a = -1$$
$$b = -4$$
$$c = 5$$
Then
$$m = 2$$
$$n = 9$$
So,
$$9 - \left(q^{2} + 2\right)^{2}$$
Factorization
[src]
/ ___\ / ___\ (q + 1)*(q - 1)*\q + I*\/ 5 /*\q - I*\/ 5 /
$$\left(q - 1\right) \left(q + 1\right) \left(q + \sqrt{5} i\right) \left(q - \sqrt{5} i\right)$$
(((q + 1)*(q - 1))*(q + i*sqrt(5)))*(q - i*sqrt(5))
Combining rational expressions
[src]
2 / 2\ 5 + q *\-4 - q /
$$q^{2} \left(- q^{2} - 4\right) + 5$$
5 + q^2*(-4 - q^2)
Combinatorics
[src]
/ 2\ -(1 + q)*(-1 + q)*\5 + q /
$$- \left(q - 1\right) \left(q + 1\right) \left(q^{2} + 5\right)$$
-(1 + q)*(-1 + q)*(5 + q^2)