Is this correct?

Question

I'm suppose to factorise this equation: x^2 - 6x - 3 (for homework).. I've worked on it and the only solution i can get is:

(x-3)^2-6 = x^2-6x-3

checked: (x-3)(x-3) -6
x^2 -3x-3x+9-6
Therefore = x^2 - 6x + 3

Is this ok and also, is there any other method of doing this?? Suggestion welcome...

Thanks and Happy Thanksgiving :-D

Answer 1

X^2 - 6X - 3 = X^2 - 6X + 9 - 9 - 3
"Cleverly add Zero - [ +9 - 9 = 0 ]"
= (X^2 - 6X + 9) - 12 "(X^2 - 6X + 9) is a Square (X-3)^2
= (X - 3)^2 - ((X^(1/2))^2
= [(X-3)-(X^(1/2))]
[(X-3)+(X^(1/2))]

Hope you're Happy now!!!!!

Answer 2

(x-3)^2 -6 = x^2 - 6x + 3 . You wanted to factorize x^2 - 6x - 3 right?

x^2-6x+3 =(x-3)^2 - ^2
= (x-3+sqrt(6))(x-3-sqrt(6))

Similarly, x^2 - 6x - 3 = (x-3)^2 - 12. Replace 6 by 12 in the previous method

Answer 3

x^2-6x-3
x^2-2*3*x+9-9-3
x^2-6x+9-12
(x-3)^2-12 is the correct solution

Answer 4

x^2-6x-3
x^2-2*3*x+9-9-3
x^2-6x+9-12
(x-3)^2-12 is the correct solution

Answer 5

IIRC, the roots of a quadratic are given by

(-b +/- sqrt(b^2-4ac))/2a

Here, a=1, b=-6 and c=-3

So the roots would be (6+/- sqrt(36-4*1*(-3)))/2 or
3+/- sqrt(48) or 3+4(sqrt(3)) and 3-4sqrt(3).

So it doesn't factor into whole number terms.

If it WERE factorable, c would have factors that sum to b (because (x+n)(x+m)=x^2+(n+m)x +nm.

The only factors of 3 are 1 and 3, but no combination of +/- 1 and 3 (1+3,1-3,3-1,-3-1) sums to -6.

In order to be factorable, assuming a and c are correct, b would have to be one of 4,-2,2, or -4.

Answer 6

two sentence from you are opposite to each other, please edit the question?

I'm suppose to factorise this equation: x^2 - 6x - 3 (for homework)..
AND
Therefore = x^2 - 6x + 3


so which one do you want, +3 or -3 at the end?

you can try ABC formula, check the link, OK.

Answer 7

It can be written in this way
(x-(3+2*sqrt(12)) * (x-(3-2*sqrt(12))