Solve and find the integer of 4x^2-19x-5
1 answer:
(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)
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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