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Simple Math Practice Problems


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last samurai is correct.

Explanation

x^4+x^2+1=

x^4+2x^2+1-x^2= (add and subtract x^2)

(x^2+1)^2-x^2= ((a+:D^2=a^2+2ab+b^2)

(x^2+x+1)(x^2-x+1) (a^2-b^2=(a+:D(a-:D)

In case anyone wants to know how to do my other one,

A polynomial of degree 4, in the form x^4+ax^3+bx^2+cx+d=0 (a,b, and c are all real numbers), has roots -4+3i and 9+4i. What is the value of a?

For a polynomial to have real coeficcients and still have complex roots, each complex root must be paired with its conjugate. So, if two roots are -4+3i and 9+4i, the other two must be -4-3i and 9-4i.

In any polynomial in the form,

ax^n+bx^(n-1)+cx^(n-2)+dx^(n-3)+ex^(n-4)...

-b/a is always the sum of the roots of the equation.

So, -a/1=-a= (-4+3i)+(9+4i)+(-4-3i)+(9-4i)=10

So, a = -10

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Akya - Oh well, if nobody else is going to guess:

f(x,y) = x^2 + 4xy + sin y + y^2

(df/dx)y = 2x + 4y

(df/dy)x = 4x + cos y + 2y

yep got it right...^^ sorry I didn't answer immediatly :banana:

how about this :

Find a polar equation for the circle x^2+(y-3)^2=9

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Deep breath... and:

A = (2+sqrt(5))^(1/3) + (2-sqrt(5))^(1/3)

A^3 = (2+sqrt(5)) + 3.((2+sqrt(5))(9-4.sqrt(5)))^(1/3) + 3.((2-sqrt(5))(9+4.sqrt(5)))^(1/3) + (2-sqrt(5))

= 4 - 3.[ (2+sqrt(5))^(1/3) + (2-sqrt(5))^(1/3) ]

= 4 - 3.A

A^3 + 3.A - 4 = 0

(A-1)(A^2+A+4) = 0

A = 1. I'm not going to bother with the imaginary roots.

Good enough? :D

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