factorise:2a³x²-5a²x-12a
Answers
30x^2*y-87xy+30y=y*(30x^2-87x+30).
30x^2-87x+30=0 solutions are 2/5 and 5/2 => 30x^2-87x+30=(x-5/2)(x-2/5)
so 30x^2y-87xy+30y=y(x-2/5)(x-5/2);
9x^2-5x-10=0 solutions are (5-sqrt(385))/18 and (5+sqrt(385))/18
therefore 9x^2-5x-10=(x-(5-sqrt(385))/18)(x-(5+sqrt(385))/18);
x^6-4x^2=x^2(x^4-4)
x^4-4=(x^2+2)(x^2-2) // a^2-b^2=(a+b)(a-b)
When working with only real roots, it's usual to leave x^2+2 as it is, since it has complex roots, but if you want you can factor it too. If you factor it it will be:
x^2+2=0, x^2=-2, solutions are i*sqrt(2) and -i*sqrt(2)
therefore: x^2+2=(x-i*sqrt(2))(x+i*sqrt(2))
x^2-2=0 has solutions sqrt(2) and -sqrt(2) therefore:
x^2-2=(x-sqrt(2))(x+sqrt(2)) //note that this is what you would get if you used a^2-b^2 formula given above
Finally, we get:
x^6-4x^2=x^2(x^2+2)(x-sqrt(2))(x+sqrt(2));
Some important conclusions:
(If the biggest power of variable in polynomial is n, we will say that that is n-th degree polynomial.)
N-th degree polynomial has N complex roots (some may appear more than once, such as x=0 in the last example, appearing as second degree root, and that is counted as two roots).
For every complex root there is it's conjugate pair, so if you have root that is a+b*i you will also have a-b*i as a root too. (This can be useful when there are simple complex roots such as i, then you already know the other one and you can divide the polynomial with (x-i)(x+i)=(x^2+1).
If x1 is a root of polynomial P (meaning that when you put P=0, x1 will be the solution) then the polynomial P is divisible by (x-x1). (This is what I did in all of the examples).
30x^2*y-87xy+30y=y*(30x^2-87x+30).
30x^2-87x+30=0 solutions are 2/5 and 5/2 => 30x^2-87x+30=(x-5/2)(x-2/5)
so 30x^2y-87xy+30y=y(x-2/5)(x-5/2);
9x^2-5x-10=0 solutions are (5-sqrt(385))/18 and (5+sqrt(385))/18
therefore 9x^2-5x-10=(x-(5-sqrt(385))/18)(x-(5+sqrt(385))/18);
x^6-4x^2=x^2(x^4-4)
x^4-4=(x^2+2)(x^2-2) // a^2-b^2=(a+b)(a-b)
When working with only real roots, it's usual to leave x^2+2 as it is, since it has complex roots, but if you want you can factor it too. If you factor it it will be:
x^2+2=0, x^2=-2, solutions are i*sqrt(2) and -i*sqrt(2)
therefore: x^2+2=(x-i*sqrt(2))(x+i*sqrt(2))
x^2-2=0 has solutions sqrt(2) and -sqrt(2) therefore:
x^2-2=(x-sqrt(2))(x+sqrt(2)) //note that this is what you would get if you used a^2-b^2 formula given above
Finally, we get:
x^6-4x^2=x^2(x^2+2)(x-sqrt(2))(x+sqrt(2));
Some important conclusions:
(If the biggest power of variable in polynomial is n, we will say that that is n-th degree polynomial.)
N-th degree polynomial has N complex roots (some may appear more than once, such as x=0 in the last example, appearing as second degree root, and that is counted as two roots).
For every complex root there is it's conjugate pair, so if you have root that is a+b*i you will also have a-b*i as a root too. (This can be useful when there are simple complex roots such as i, then you already know the other one and you can divide the polynomial with (x-i)(x+i)=(x^2+1).
If x1 is a root of polynomial P (meaning that when you put P=0, x1 will be the solution) then the polynomial P is divisible by (x-x1). (This is what I did in all of the examples).
Step-by-step explanation:
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