no.
(x^2+3x+4). What multiplys to 4 and adds to three? Nothing. There are no factors. If the 4 were negative, then 4,-1 would be the factors (x+4)(x-1), but that is not the case.
You need to learn how to factor, it is not a guessing game.
Well I'm curently taking this class as a sophmore and have taken geometry freshmen year of high school and also I am taking physics at the same time and in math class when I'm asked to factor i go absolutley nuts because guessing numbers is just no my thing... I need solve for X rearange formulas and so forth...
with that said is this legal?
X^2 + 3X + 4
factor that...
(X^2 + 3X + 4)^(1/2)
factored form
(X + (3X)^(1/2) + 2)^2
I believe this is perfectly acceptable for factoring but I'm not sure...
my textbook says that the agreement for completely factored from is..
A polynomial is completely factored when it is written as a product of tow or more polynomilas with integers for their coefficients.
So I believe I can leave that as a factored form correct?
4 answers
The square root of a polynomial is not the sum of the square roots of each term. It just doesn't work that way.
If you are trying to factor
x^2 + 3x + 4, there are no simple monomial factors with integer constants. The roots are complex,
[-3 +/- sqrt7i]/2 and -1.5 +/- 2.6458i
The factors are
(x+1.5-2.6458i)(x+1.5+2.6458i)
where i is the square root of -1
If you are trying to factor
x^2 + 3x + 4, there are no simple monomial factors with integer constants. The roots are complex,
[-3 +/- sqrt7i]/2 and -1.5 +/- 2.6458i
The factors are
(x+1.5-2.6458i)(x+1.5+2.6458i)
where i is the square root of -1
The roots are complex,
[-3 +/- isqrt7]/2 = -1.5 +/- 1.3228i
The factors are
(x+1.5-1.3228i)(x+1.5+1.3228i)
where i is the square root of -1
[-3 +/- isqrt7]/2 = -1.5 +/- 1.3228i
The factors are
(x+1.5-1.3228i)(x+1.5+1.3228i)
where i is the square root of -1
2pie16