How do you factor #2d^2-7d-15#?

2 Answers
Nov 17, 2015

Answer:

#(2d+3)(d-5)#.

Explanation:

This may be factorized as a trinomial to obtain the factors

#(2d+3)(d-5)#.

To obtain this you look at the coefficient of the first term, 2, and see what its factors are. There is only 1 option, #1xx2#.

Then look at the coefficient of the last term, 15, and there are a 2 options here namely :
#1xx15 or 3xx5#.

You now need to combine the options such that the sum of the factors when cross-multiplied give you the middle term of #-7#.

The only option is #((2,3),(1,-5))# since #1xx3-2xx-5=-7#

and so therefore the factors are the rows of the matrix shown.

Nov 17, 2015

Answer:

#(2d+3)(dāˆ’5)#

Explanation:

Factor #2d^2#āˆ’#7d#āˆ’#15# ?

To factor a quadratic function with a leading coefficient greater than 1, do the following steps:

  • Multiply the leading coefficient and the constant:

#2 * #-#15 = #-#30#

  • What two numbers would multiply to give you -#30# and add to give you -#7# ?

#3 * #-#10 = #-#30#
#3 + #-#10 = #-#7#

So, +#3# & -#7#.

  • Since we multiplied the constant by 2, divide the numbers you got above by 2:

#+3/2 & -10/2#

#-10/2 = #-#5# so, one of the factors is #(d-5)#
but #+3/2# leaves you with an improper fraction, right? (Yes.)
So, instead of #(d+3)# we're going to have #(2d+3)#.

  • Final factors:

#(2d+3)(d-5)#