9x^2-6x+1

I don't understand how to factor the expression.

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To factor 9x^2-6x+1.

We regroup the middle term ,-6x of the given expresion as below:

9x^2 **-3x -3x**+1

=3x(x-1) -1(3x-1)

=(3x-1)^2 .

Aliter:

We know that a^2+2ab+b^2 = (a+b)^2 is an identiy. Use this to write the given expression to bring to the form, (a+b)^2.

9x^2 -6x +1. Term by term we can write this like:

=(3x)^2 + 2(3x)(-1) + (-1)^2. a = 3x and b = -1.Threfore,

=[3x+(-1)]^2

(3x-1)^2.

Factoring an expression means representing the expression as multiple of two or more terms. In a quadratic equation, that is an equation in which one of the term contains square of the variable (say x), and no term which has higher power than square. There are always maximum two factors containing the term x. Thus the factors of the expression of the type:

ax^2 + bx + c

will be of the form:

(x - A)(x -B)*C

Where A, B and C are constant numbers.

There are formulae to calculate the values of A, B and C from the values of a, b and c, without finding out the factors. But in many cases it is possible to find out the factors with a little bit of ingenuity or a little bit of trial and error. To illustrate how this can be done we will find out factors of the given expression in two different ways.

Method I:

If we examine the expression 9x^2 - 6x + 1 carefully, we can see that it conforms to the form p^2 - 2pq - q^^2, which is equal to (p - q)^2. Thus we can straight away we can find out the factors of the expression as:

(3x - 1)^2 = (x - 1/3)(x -1/3)*9

Method II

In this method we need term bx of the expression ax^2 + bx + c in two components b'x + b"x so that we can find a common factor in the terms (ax^ + b'x) and (b"x + c).

Thus we can represent the given expression as:

9x^2 - 3x - 3x + 1

= 3x*(3x - 1) -1*(3x - 1)

= (3x - 1)(3x - 1)

= (x - 1/3)(x - 1/3)*9

To factor trinomials (polynomials with 3 terms) do the following:

1. Check to see if you can divide all three terms by the same number or variable. In this example you cannot.

2. Since the first term 9x^2 and the last term 1 are perfect squares, the polynomial might be a perfect square.

3. To factor a perfect square take the square root of the first term, the sign of the second term and the square root of the last term, put them in parantheses and square the quantity:

(3x - 1)^2

To verify that this is a perfect square multiply the two terms inside the parantheses and double

3x times -1 = -3x doubled is -6x if this is the middle term of the given poly. It is factored correctly.

Or you could write out both factors

(3x -1) (3x - 1) and use the FOIL method to make sure you get the given polynomial

9x^2-6x+1

a=9 b = -6 c = 1

just multiply a by c

9 x 1 = 9

then find factors of 9 that add up to b (-6) which would be -3 and -3 because:

-3 + -3 = -6 and 3 x 3 = 9

now just plug the factors as b

9x^2 - 3x - 3x + 1

group the problem

(9x^2 - 3x ) ( - 3x + 1)

factor out the greatest common factors:

3x (3x - 1) -1 ( 3x - 1)

now put the numbers outside of the parenthesis together

(3x - 1) (3x-1)

(that's the answer unless you are trying to find the roots. To find the roots:

set them equal to 0) since the numbers are the same we just need to do this once:

3x - 1 = 0

solve

3x - 1 =0

+1 +1

3x = 1

x = 1/3

the root is `+-1/3`

The expression 9x^2-6x+1 has to be factored.

For a quadratic expression ax^2 + bx + c, to factor it, write b as a sum of two numbers n1 and n2 such that n1*n2 = a*c and b = n1+n2

9x^2-6x+1

= 9x^2 - 3x - 3x + 1

factor the common term 3x

= 3x(3x - 1) - 1(3x - 1)

= (3x - 1)^2

The factorized form of 9x^2-6x+1 is (3x - 1)^2

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