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)
You might be interested in
Answer:
Option A is correct
and
Step-by-step explanation:
Given the functions:
then;
combine them to create an arithmetic sequence i.e,
⇒ where n is the number of terms
To find the 31st term.
Substitute n =31 we have;
⇒
Therefore, we have an arithmetic sequence,
and