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20 November, 09:26

Solve and find the integer of 4x^2-19x-5

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  1. 20 November, 12:32
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    (x-5) (4x+1)

    4x^2 times - 5 is - 20 the factors of - 20 are - 1x20, 1x-20 ... but to make the - 19x you would add together the 1 and - 20 because 1-20 is - 19.

    This then splits off to factorise 4x^2-20x, making this side 4x (1x-5) carry the brackets over to make it 1 (1x-5) joining the brackets together to make (4x+1) (x-5)
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