Can #y=2x^2 – 3x – 5# be factored? If so what are the factors ?

3 Answers
Apr 9, 2018

Answer:

#x=5/2,-1#

Explanation:

#2x^2-3x-5#

=#(2x-5)(x+1)#

#x=5/2,-1#

Apr 9, 2018

Answer:

#(x+1)(2x-5)#

Explanation:

#"for a quadratic in "color(blue)"standard form"#

#•color(white)(x)y=ax^2+bx+c color(white)(x);a!=0#

#"consider the factors of the product ac which sum to b"#

#"here "a=2,b=-3" and "c=-5#

#rArrac=2xx-5=-10" and - 5, + 2 sum to - 3"#

#color(blue)"split the middle term using these factors"#

#rArry=2x^2+2x-5x-5larrcolor(blue)"factor by grouping"#

#color(white)(y)=color(red)(2x)(x+1)color(red)(-5)(x+1)#

#"take out a "color(blue)"common factor "(x+1)#

#=(x+1)(color(red)(2x-5))#

#rArr2x^2-3x-5=(x+1)(2x-5)#

Apr 9, 2018

Answer:

#y=(2x-5)(x+1)#

Explanation:

Can the trinomial be factored? #y = ax^2 +bx+c#

Test it using #" "b^2 -4ac" "# first.
If the answer is a perfect square then you can find factors. This means you do not waste time by trying to find factors if there are none!

#(-3)^2 -4(2)(-5) = 9+40 =49" "larr = 7^2#

Find factors of #2 and 5# whose products differ by #3#

Both #2 and 5# are prime numbers so there are only two combinations to try:

#(2xx5) - (1xx1) = 10-1=9#

#(1xx5)-(2xx1) =5-2=3#

#y=(2x-5)(x+1)#