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

Answer 1

Hazel Anglin

#2d^2+7d-15 = (d+5)(2d-3)#

For this one, I used the quadratic formula.

#a=2#, #b=7#, #c=-15#

#Delta = b^2-4ac = 49+120 = 169 = 13^2#

So #2d^2+7d-15 = 0# has roots:

#d = (-b+-sqrt(Delta))/(2a)#

#=(-7+-13)/4#

That is #d = -5# or #d = 3/2#

Hence factors #(d+5)# and #(2d-3)#, respectively.

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Answer 2

Eli Assouline

To factor the quadratic expression 2d^2 + 7d - 15, you need to find two numbers that multiply to give you the product of the coefficient of the squared term (2) multiplied by the constant term (-15), and add up to the coefficient of the linear term (7).

The numbers are 5 and -3:

2 * (-15) = -30 5 + (-3) = 2

Then, you can rewrite the expression as:

2d^2 + 10d - 3d - 15

Now, you can factor by grouping:

2d^2 + 10d - 3d - 15 = 2d(d + 5) - 3(d + 5)

Factor out the common factor:

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

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.