Factorise: x squared - 2x + c?

Question

In the form (x +/- a)(x +/- b) stating the value of a, b and c, where c is an integer.

Factorise: x squared - 2x + c?
In the form (x +/- a)(x +/- b) stating the value of a, b and c, where c is an integer. Both a and b have to be different i.e the answer cannot be (x -1)(x -1)

Answer 1

Note: if c=1, then this is a perfect square:

x^2 - 2x + 1 = (x-1)^2

Now, more generally, if c is some integer, then there are two roots, x1 and x2, that are expressed as:

x1 = (-b - sqrt( b^2 - 4*a*c ) )/(2*a)

and

x2 = (-b - sqrt( b^2 - 4*a*c ) )/(2*a)

where, in your case, a=1, b=-2, and c=c. (in other words, the polynomial is a*x^2 + b*x + c) The notation sqrt(blah) represents the square root of blah.

That is:

x1 = (2 - sqrt( 4 - 4c ) )/2
x1 = (2 - 2*sqrt(1-c) )/2
x1 = 1 - sqrt(1-c)

and

x2 = (2 + sqrt( 4 - 4c ) )/2
x2 = (2 + 2*sqrt(1-c) )/2
x2 = 1 + sqrt(1-c)

And so you want:

(x - (1 - sqrt(1-c)))(x - (1 + sqrt(1-c)))

in other words:

(x - 1 + sqrt(1-c))(x - 1 - sqrt(1-c))

and that's your factorization. Again, note that if c=1, then this reduces to (x-1)(x-1) which is (x-1)^2, a perfect square.

Answer 2

First, a comment: Factorise in the UK is the same
as our verb "factor" in the USA.
Actually there are infinitely many c that will yield
a factorisation over the integers. Let's find them.
To have such a factorisation, the discriminant
of the quadratic must be a square(which means
it must certainly be nonnegative).
So 4-4c is a square, i.e., 1-c is a square, say y^2.
If c >= 0 then c must be 0 or 1.
But if c < 0, then -c = y^2-1, so any value
of -c one less than a square will yeld such a factorisation.
Examples:
x^2 - 2x -3 = (x-3)(x+1)
x^2 -2x -8 = (x-4)(x+2)
x^2 -2x -15 = (x-5)(x+3)
x^2 -2x -24 = (x-6)(x+4)

Answer 3

To FACTOR (not factorise) this you need to know what c is
c would have to be 1 in order to get -2x in the middle

x^2 - 2x + 1
(x-1)(x-1)

Answer 4

x^2-2x+c
if we have an equation
ax^2+bx+c then its factors are
x={-b+-sqrt[b^2-4ac}}/2a
please observe there are two roots
one with + sign and other with- sign
we use this for the given expression
x={2+-sqrt[4-4c]}/2
=2{1+-sqrt[1-c]}/2
=1+-sqrt[1-c]
the factors are
[x-1-sqrt(1-c)][x-1+sqrt(1-c)]

Answer 5

"completing the square" gives,
x^2 - 2*x + c =
x^2 - 2*x + 1 - (1-c) =
(x-1)^2 - sqrt(1-c))^2 =
(x-1+sqrt(1-c)) * (x-1-sqrt(1-c))

assuming form (x+a) * (x+b),
a = sqrt(1-c) - 1
b = - (1+sqrt(1-c))

Answer 6

The roots could be gained via using the formula {-b +/- sqrt(b^2-4ac)}/ 2a = {2 +/- sqrt(4-4c)}/2 = a million +/- sqrt(4-4c)/2 = a million +/- sqrt(a million-c) ; c<=a million => {x - a million - sqrt(a million-c)} {x - a million + sqrt(a million-c)}

Answer 7

A little rusty here:

(x-1)(x-1)
a=-1
b=-1
c=1

Answer 8

x^2-2x+c
(x - a ) ( x -b )

Where -a + -b = -2 and -a x -b = c

Answer 9

x=[2+/-rt(4-4c)]/2
[2+/-2rt(1-c)]/2
[1+rt(1-c)] and [1-rt(1-c)]
so the factors would be
[x+{1+rt(1-c)}][x-{1+rt(1-c]}]

Answer 10

(x-c)(x-c)

since 'c' is just a constant, it can be altered in any way.

(x-c)^2 should be sufficient