How do you factor $4 x^{2} - 5 x + 1 = 0$ $4x^2 - 5x + 1 = 0$ ?

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How do you factor $4 x^{2} - 5 x + 1 = 0$ $4x^2 - 5x + 1 = 0$ ?

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Answer 1 · 2017-02-20T22:43:28

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

Explanation:

Given:

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

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Method 1

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

$4 - 5 + 1 = 0$ $4-5+1 = 0$

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

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

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Method 2

Note that the prime factorisation of $451$ $451$ is:

$451 = 11 \cdot 41$ $451 = 11*41$

and this multiplication involves no carrying of digits.

Hence we find:

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

and:

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

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Method 3

Note that:

$(a x - 1) (b x - 1) = a b x^{2} - (a + b) x + 1$ $(ax-1)(bx-1) = abx^2-(a+b)x+1$

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

Hence we find:

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

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How do you factor $4 x^{2} - 5 x + 1 = 0$ $4x^2 - 5x + 1 = 0$ ?

1 Answer

Explanation:

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How do you factor $4 x^{2} - 5 x + 1 = 0$ $4x^2 - 5x + 1 = 0$ ?