28 ( x2, x3 and x4)
You sure can, here is an example I made up to show you: x4+x2= x2(x2+x) When you factor it out it should look like this because if you multiply it back up (where the exponents will multiply not add) it will equal the primary equation.
x5-1 = (x - 1)(x4 + x3 + x2 + 1)
(x4 - 16) = (x2 + 4) (x2 - 4) = (x2 + 4) (x + 2) (x - 2)
I assume the x4 is x to the 4th power, and x2 is x², or x squared. The two factors would be: (x + 2)(x - 2)(x² + 3) There are 4 solutions for x: x = 2 x = -2 x = √-3 x = -√-3
x4 + 7x2 - 60 = x4 + 12x2 - 5x2 - 60 = x2(x2 + 12) - 5(x2 +12) = (x2 - 5)(x2 +12)
x4 +x2 =x2 (x2+1)
1 - x4 = (1 - x2)(1 + x2) = (1 - x)(1 + x)(1 + x2) (difference of squares)
(x2 - 3)(x2 + 3)(x4 + 9)
-6x3 + 4x7 - 0.8x2 + x4 - 5x5 = 4x7 - 5x5 + x4 - 6x3 - 0.8x2 = x2 (4x5 - 5x3 + x2 - 6x - 0.8)
If there is a remainder then it's not a factor otherwise yes
(xn+2-1)/(x2-1)ExplanationLet Y=1+x2+x4+...+xn. Now notice that:Y=1+x2+x4+...+xn=x2(1+x2+x4+...+xn-2)+1Y+xn+2=x2(1+x2+x4+...+xn-2+xn)+1Y+xn+2=x2*Y+1Y+xn+2-x2*Y=1Y-x2*Y=1-xn+2Y(1-x2)=1-xn+2Y=(1-xn+2)/(1-x2)=(xn+2-1)/(x2-1)
(x4 - 2x3 + 2x2 + x + 4) / (x2 + x + 1)You can work this out using long division:x2 - 3x + 4___________________________x2 + x + 1 ) x4 - 2x3 + 2x2 + x + 4x4 + x3 + x2-3x3 + x2 + x-3x3 - 3x2 - 3x4x2 + 4x + 44x2 + 4x + 40R∴ x4 - 2x3 + 2x2 + x + 4 = (x2 + x + 1)(x2 - 3x + 4)
-32
x(x4+x3+x2+x+1)
-x2(x + 10)(x - 10)
x4 - 1.We can not "solve" this as we have not been told the value of x. However, we can simplify this expression:We have an x and a minus x here which will cancel out. Likewise the x2 and x3 will cancel out with the -x2 and -x3 respectively. This therefore leaves us with just x4 - 1.