How do you factor completely $4x^2- 13x + 10$ ?

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Answer 1 · 2015-11-20T09:56:55

$(x-2)(4x-5)$

$4x^2-13x+10$

Multiply co-efficent of $x^2$ with the constant term $10$

It is $40$
The co-efficeint of $x=-13$

You have to find two numbers the product of those two numbers must be equal to $40$ and the sum of the two numbers must be equal to $-13$

They are $-8, -5$

Now rewrite the expression as -

$4x^2 -5x-8x+10$

Factor it -

$x(4x-5)-2(4x-5)$

$(x-2)(4x-5)$

How do you factor completely 4x^2- 13x + 104x^2- 13x + 10?