How do I factor 3z^2 - 14z - 24?

Answer 1

Hi,

I' m going to explain factoring the way that I teach it, which is probably different from what you've done. Try to follow the steps. This method will work on any trinomial.

3z^2 - 14z - 24 Look for a GCF None this time. If there was one, factor it out. Then temporarily start both
...........................parentheses with first number and variable.
(3z.......)(3z..........) First sign goes in first parentheses.
(3z..-....)(3z..........) Product of signs goes in 2nd parentheses.
(3z..-....)(3z...+.....) <== plus is because neg x neg = positive

Now multiply your first and third numbers together. Ignore their signs - you've already done them. 3 x 24 = 72 So, out to the side list pairs of factors of 72.

72
------
1, 72
2, 36
3, 24
4, 18
6, 12
8, 9

Now you want to pick which factors go in your parentheses, using these rules:

If the signs you put on your parentheses are the SAME, find the factors that ADD up to the middle number in the problem.
If the signs you put on your parentheses are the DIFFERENT, find the factors that SUBTRACT to the middle number in the problem. (Note you look at signs in the parentheses, not in the problem.)

(3z..-....)(3z...+.....) Your signs are different, so you want to subtract factors to get 14. Those factors are 4 and 18. When you put the numbers in the parentheses, the bigger number is pushy - it always goes first. So your factors now are:
(3z..-..18)(3z.+.4.)

Now you have to reduce either or both parentheses by dividing each parentheses' terms by the largest possible divisor. In our problem, the first parentheses is divisible by 3 but the second parentheses does not reduce.

(3z..-..18)(3z.+.4.)
-------------
.......3 This reduces to your final factors of

(z - 6)(3z + 4) <== This is your answer.

Here are some more examples:

3y^2 - 13y - 10 Look for a GCF None this time. If there was one, factor it out. Then temporarily start both
...........................par... with first number and variable.
(3y.......)(3y..........) First sign goes in first parentheses.
(3y..-....)(3y..........) Product of signs goes in 2nd parentheses.
(3y..-....)(3y...+.....) <== plus is because neg x neg = positive

Now multiply your first and third numbers together. Ignore their signs - you've already done them. 3 x 10 = 30 So, out to the side list pairs of factors of 30.

30
------
1, 30
2, 15
3, 10
5, 6

Now you want to pick which factors go in your parentheses, using these rules:

If the signs you put on your parentheses are the SAME, find the factors that ADD up to the middle number in the problem.
If the signs you put on your parentheses are the DIFFERENT, find the factors that SUBTRACT to the middle number in the problem. (Note you look at signs in the parentheses, not in the problem.)

(3y..-....)(3y...+.....) Your signs are different, so you want to subtract factors to get 13. Those factors are 2 and 15. ( Notice that 3 and 10 would have added to 13, so you HAD to know to subtract this time.) When you put the numbers in the parentheses, the bigger number is pushy - it always goes first. So your factors now are:
(3y..-..15)(3y.+.2.)

Now you have to reduce either or both parentheses by dividing each parentheses' terms by the largest possible divisor. In our problem, the first parentheses is divisible by 3 but the second parentheses does not reduce.

(3y..-..15)(3y.+.2.)
-------------
.......3 This reduces to your final factors of

(y - 5)(3y + 2)

NEXT PROBLEM !!

2x^2 +15x + 7 Look for a GCF None this time. If there was one, factor it out. Then temporarily start both
...........................par... with first number and variable.
(2x.......)(2x..........) First sign goes in first parentheses.
(2x..+....)(2x..........) Product of signs goes in 2nd parentheses.
(2x..+....)(2x...+.....) <== plus is because pos x pos = positive

Now multiply your first and third numbers together. Ignore their signs - you've already done them. 2 x 7 = 14 So, out to the side list pairs of factors of 14.

14
------
1, 14
2, 7

Now you want to pick which factors go in your parentheses, using these rules:

If the signs you put on your parentheses are the SAME, find the factors that ADD up to the middle number in the problem.
If the signs you put on your parentheses are the DIFFERENT, find the factors that SUBTRACT to the middle number in the problem. (Note you look at signs in the parentheses, not in the problem.)

(2x..+....)(2x...+.....) Your signs are the same, so you want to add factors to get 15. Those factors are 1 and 14. When you put the numbers in the parentheses, the bigger number is pushy - it always goes first. So your factors now are:
(2x.+.14)(2x.+.1)

Now you have to reduce either or both parentheses by dividing each parentheses' terms by the largest possible divisor. In our problem, the first parentheses is divisible by 2 but the second parentheses does not reduce.

(2x.+.14)(2x.+.1)
-------------
.......2 This reduces to your final factors of

(x + 7)(2x + 1)


I hope that helps you!! :-)

Answer 2

3z^2 - 14z - 24
= 3 z^2 + 4 z - 18z - 24
= z(3z+4) - 6(3z + 4)
= (z-6)(3z+4)

Answer 3

given : 3z^2 - 14z -24
= 3z^2 - 18z + 4z - 24
= 3z(z-6) + 4(z-6)
= (3z+4)(z-6)

Answer 4

guess and check
since you know the factors of 3

it would be

(3z )(z )


then you would guess and check on the factors of 24



so the answer would be

(3z-4)(z+6)


and then to make sure you would foil it

its 3z^2 - 14z - 24


tada :)

Answer 5

(3z+4)(z-6)=0 z=6 or z=-4/3

Answer 6

hi this question is too easy to solve ok
first we must be arrange by z
1)3z^2-14z-24=0
z(3z-14)=24
z=24 or
3z-14=24------->3z=24+14-------->3z=28
then z=28divided by 28
i hope help you

Answer 7

3z^2 - 14z - 24

z = (-(-14) ± sqrt((-14)^2 - 4(3)(-24)))/(2(3))
z = (14 ± sqrt(196 + 288))/6
z = (14 ± sqrt(484))/6
z = (14 ± 22)/6
x = (-8/6) or (36/6)
x = (-4/3) or 6

ANS : (x - 6)(3x + 4)

Answer 8

(z-6)(z+4/3)

Answer 9

(z-6)(3z+4)

Answer 10

Take a look at this page (it looks better on this page)

http://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/Polynomials-and-rational-expressions.faq.question.78901.html

If you need more help with factoring, check out this solver (it will guide you step by step)

http://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/quadratic-factoring.solver