Factor the following completely Im not sure how to do this please show work I will give medals

1. 3x^2 + 8x – 16
2. c^2 – 121
3. p^2 + 22p + 40

Question

Factor the following completely Im not sure how to do this please show work  I will give medals
 
1.  3x^2 + 8x – 16 
2.  c^2 – 121 
3.  p^2 + 22p + 40

luigi0210 · Answer

Well 2 is the difference of two perfect squares.

anonymous · Answer

3*-16=-48
12-4=8
12*-4=-48

anonymous · Answer

is that the answer for the first one?

anonymous · Answer

yes,write 8x=12x-4x and make factors.

anonymous · Answer

im confused

whpalmer4 · Answer

Split the middle term:
$$3x^2 + 8x -16 = 3x^2 + 12x - 4x -16$$Now rewrite as two groups:
$$(3x^2+12x) - (4x+16)$$(notice the sign change on the 16 as I factored out the minus sign!)
Now factor each group separately:
$$3x(x+4) - 4(x+4)$$Notice that each of those expressions has a common factor?
$$(x+4)(3x-4)$$
Check the work:
$$(x+4)(3x-4) = x(3x-4) + 4(3x-4) = 3x^2 - 4x + 12x -16 $$$$\qquad= 3x^2 +8x -16\checkmark$$

anonymous · Answer

thanks. what about the other two?

whpalmer4 · Answer

You come up with the coefficients to split it by multiplying the coefficient of the 1st term with the coefficient of the 3rd term: 3*-16 = -48.  Now find a pair of factors of -48 that add to 8: 12 and -4 are that pair.

You can use the same procedure on the last one:
$$p^2 + 22p + 40$$1*40 = 40.  Factors of 40 that add to 22 are 2 and 20.  I'll let you do the rest and check your work.

@Luigi0210 already told you how to do the 2nd one.  $$a^2-b^2 = (a+b)(a-b)$$