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

Feb 20, 2017

$\left(4 x - 1\right) \left(x - 1\right) = 0$

#### Explanation:

Given:

$4 {x}^{2} - 5 x + 1 = 0$

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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 $\left(x - 1\right)$ a factor:

$0 = 4 {x}^{2} - 5 x + 1 = \left(x - 1\right) \left(4 x - 1\right)$

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Method 2

Note that the prime factorisation of $451$ is:

$451 = 11 \cdot 41$

and this multiplication involves no carrying of digits.

Hence we find:

$4 {x}^{2} + 5 x + 1 = \left(x + 1\right) \left(4 x + 1\right)$

and:

$4 {x}^{2} - 5 x + 1 = \left(x - 1\right) \left(4 x - 1\right)$

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Method 3

Note that:

$\left(a x - 1\right) \left(b x - 1\right) = a b {x}^{2} - \left(a + b\right) x + 1$

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:

$\left(4 x - 1\right) \left(x - 1\right) = 4 {x}^{2} - 5 x + 1$