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How do you factor #x^2+3x+4x+12# by grouping?

Answer 1

Abigail Allen

#(x + 3) (x + 4)#

#x^2 + 3x + 4x + 12#

Grouping;

#(x^2 + 3x) (+4x + 12)#

Factoring;

#x(x + 3) +4(x + 3)#

#(x + 3) (x + 4)#

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Answer 2

Eva Abramson

#(x+3)(x+4)#

Let's look at our quadratic as two parts:

#color(steelblue)(x^2+3x)+color(purple)(4x+12)#

We see that the blue terms have an #x# in common, and the purple terms have a #4# in common, so we can factor that out to get

#color(steelblue)(x(x+3))+color(purple)(4(x+3))#

We now see that both terms have an #x+3# in common, so we can finally factor that out to get

#(x+3)(x+4)#

Hope this helps!

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Answer 3

Abigail Allen

To factor the expression (x^2 + 3x + 4x + 12) by grouping, we group the terms in pairs:

(x^2 + 3x + 4x + 12)
(= (x^2 + 3x) + (4x + 12))

Then, we factor each pair separately:

(= x(x + 3) + 4(x + 3))

Now, we can see that both terms have a common factor of (x + 3):

(= (x + 3)(x + 4))

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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.