How do you factor examples b), e) and f) below?

b) #y = 8x^2-10x-3#

e) #y = 5x^3-80x#

f) #y = 36x^2-1#

1 Answer
Jan 17, 2017

Answer:

b) #y = 8x^2-10x-3 = (4x+1)(2x-3)#

e) #y = 5x^3-80x = 5x(x-4)(x+4)#

f) #y = 36x^2-1 = (6x-1)(6x+1)#

Explanation:

Example b)

#y = 8x^2-10x-3#

This quadratic is in the form #ax^2+bx+c#, with #a=8#, #b=-10# and #c = -3#.

In order to tell whether it factors 'nicely', let us first check the discriminant:

#Delta = b^2-4ac = (-10)^2-4(8)(-3) = 100+96 = 196 = 14^2#

Since this is a perfect square, the given quadratic will factor exactly.

Let's use an AC method to find the factors:

Look for a pair of factors of #AC = 8*3 = 24# which differ by #10#.

The pair #12, 2# works.

Use this pair to split the middle term and factor by grouping:

#y = 8x^2-10x-3#

#color(white)(y) = 8x^2-12x+2x-3#

#color(white)(y) = (8x^2-12x)+(2x-3)#

#color(white)(y) = 4x(2x-3)+1(2x-3)#

#color(white)(y) = (4x+1)(2x-3)#

#color(white)()#
Example e)

#y = 5x^3-80x#

For this example, we can separate out the common factor #5x#, then use the difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

with #a=x# and #b=4# as follows:

#y = 5x^3-80x#

#color(white)(y) = 5x(x^2-16)#

#color(white)(y) = 5x(x^2-4^2)#

#color(white)(y) = 5x(x-4)(x+4)#

#color(white)()#
Example f)

#y = 36x^2-1#

Notice that both the terms are perfect squares, so we can use the difference of squares identity again...

#y = 36x^2-1 = (6x)^2-1^2 = (6x-1)(6x+1)#