How do you factor the expression #4x^2 - 7x - 15#?

1 Answer
Mar 21, 2016

Answer:

#x= (7+-12sqrt(2))/8#

I will let you finish this off

Explanation:

Possible factors of 15 are 1 and 15 or 3 and 5
Possible factors of 4 are 1 and 4 or 2 and 2

Let try

The 15 has to be negative that means that one of the constants is positive and the other is negative

#(2x+3)(2x-5) = 4x^2-10x +6x color(red)(...... "Fail ")#

#(4x+3)(x-5) = 4x^2-20x+3xcolor(red)(.........."Fail ")#

#(4x-15)(x+1)= 4x^2-15x+4xcolor(red)(..........."Fail ")#

Looks as though we are going to need to use the formula!

#y=ax^2+bx+c" " -> " "x=(-b+-sqrt(b^2-4ac))/(2a)#

a=4; b=-7; c=-15

#=>x=(7+-sqrt((-7)^2-4(4)(-15)))/(2(4))#

#x= (7+-sqrt(288))/8#

#x= (7+-sqrt(2xx12^2))/8#

#x= (7+-12sqrt(2))/8#

I will let you finish this off
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If you are not sure what you can use to take the root of; use a factor tree. This is the one I built for #sqrt(288)#. Look for numbers that are squared.
Tony B