Question

How do you factor `4x^2 - 5x + 1 = 0`?

Answer 1

`(4x-1)(x-1) = 0`

Explanation:

Given:

`4x^2-5x+1 = 0`

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Method 1

Note that the sum of the coefficients is `0`. That is:

`4-5+1 = 0`

Hence `x=1` is a zero and `(x-1)` a factor:

`0 = 4x^2-5x+1 = (x-1)(4x-1)`

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Method 2

Note that the prime factorisation of `451` is:

`451 = 11*41`

and this multiplication involves no carrying of digits.

Hence we find:

`4x^2+5x+1 = (x+1)(4x+1)`

and:

`4x^2-5x+1 = (x-1)(4x-1)`

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Method 3

Note that:

`(ax-1)(bx-1) = abx^2-(a+b)x+1`

Since the constant term of the given example is `1`, look for a pair of factors of the leading coefficient `4` with sum matching the middle coefficient `5`. The pair `4, 1` works.

Hence we find:

`(4x-1)(x-1) = 4x^2-5x+1`