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

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Answer 1 · 2015-11-20T09:56:55

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

$4 x^{2} - 13 x + 10$ $4x^2-13x+10$

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

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

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

They are $- 8, - 5$ $-8, -5$

Now rewrite the expression as -

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

Factor it -

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

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

How do you factor completely 4x2−13x+104x^2- 13x + 10?