Factoring Trinomials help?

Question

1. 5x^2+2x-3
2. 6a^2 +5a+1
3. 8b^2+2b-3
4. 2a^2+19a-10
5. 6y^2-11y-10

And please show me how to do these problems and not just tell me the answer for a few. ANd Hurry!!!

Answer 1

I'm only going to help you with the 1st expression because, its important for you to factor the rest.

1. 5x^2+2x-3

first: multiply the 1st & 3rd term to get -15. find two numbers that give you -15 when multiplied & 2 (2nd term) when added/subtracted. the numbers are (-3 & 5). rewrite the expression with the new middle terms.

5x^2 - 3x + 5x - 3

Sec: group "like" terms & factor both sets of parenthesis.

(5x^2 + 5x) - (3x - 3)
5x(x + 1) - 3(x + 1)

Third: the inner terms have to match - combine the inner term (once) with the outer term.

(x + 1)(5x - 3) or, (5x - 3)(x + 1)

Answer 2

You must first determine what 2 numbers you will use to factor the squares i.e. (6a^2 can be 6a x a or 3a x2a) then do the same with the numbers at the end of the equation. after that you will be able to figure out which factors you found are correct by seeing which ones will give you the value in the middle part of the equation. sounds complicated but is is not.
1.5x^2+2x-3
(5x-3)(x+1)

2 6a^2+5x+1
(3a+1)(2a+1)

3 8b^2+2b-3
(4b+3)(2b-1)

4 2a^+19a-10
(2a-1)(a+10)

5 6y^2-11y-10
(3y+2)(2y-5)

Answer 3

1) 5x^2 is automatically 5x times x.
this implies that 5x^2+2x-3 = (5x+m)(x+n)
now we only need to find m and n
since the last term of the expression is -3, then m times n must be also equal to -3. if m=-3, n=1
(5x-3)(x+1)=5x^2-3x+5x-3 = 5x^2 + 2x - 3.
(note: this is already correct, but for instance if we got a different expression, we can still use m=3, n= -1.)

2) 6a^2 + 5a + 1
in this case, m and n are obviously both equal to 1. ( because the second term of the expression is positive. if it was negative, then m=n=-1)
if we have (3a+1)(2a+1), since 3 times 2 is 6,
(3a+1)(2a+1)=6a^2+5a+1

answers to the rest of the problems (do the same thing):
3) (2b-1)(4b+3)
4) (2a-1)(a+10)
5) (3y+2)(2y-5)
hope this helps!;)

Answer 4

1) Factors of 5 = 5 and 1. Factors of 3 = 3 and 1. Find a pair that works : 5(1) + 1(-3) : so it is (5x - 3)(x + 1)

2) Factors that work : 6(1) + 1(-1) so it is (6a -1)(a + 1)

3) Factors 8 , 1 ; 4 , 2 and 1 , 3 : Try 2(3) + 4(-1) so its (2b - 1)(4b + 3)

4) Factors 2 , 1 and 1 , 10 ; 2 , 5 : Try 2(10) + 1(-1) so its (2a - 1)(a + 10)

5) Factors 1 , 6 ; 2 , 3 and 1 ,10 ; 2 , 5 : Try 3(-5) + 2(2) so its (3y + 2)(2y - 5)

Answer 5

5x^2+2x-3 start by multipling the outside coefficients (5*3)=15 then find the factors of 15 + or - when added together equal the middle coefficient (2)

factors of 15= 5,3 and 1,15
there is no combination of 1 and 15 that =2, but adding 5 and subtracting 3 give us 2. now, put them into the form
(ax+b)(cx+d) where a,b,c, and d are the factors of 15 (5,3,1,1)

start with a
(5x+b)(cx+d) if we make a 5, then we need to make c=1 in order to have 5x^2, right?
(5x+b)(1x+d) now, the only 2 numbers to place are 3,1 and d=3 because we need to have -3 leaving b=-1 3+(-1)=2
(5x-1)(x+3)

Answer 6

OK, let's do the first one: 5x^2+2x-3

First, look at the 5 in front of the x^2. The only way to factor that is 5*1 (or -5*-1, but that can be ignored for the first term).

So our answer will be of the form (5x ............)*(x ............)

Next look at the -3 at the end. There are two ways to get that:
-3*1, or 3*-1. This gives us four possibilities:

(5x - 3)*(x + 1)
(5x + 1)*(x - 3)
(5x + 3)*(x - 1)
(5x - 1)*(x + 3)

Then FOIL these out to see what the middle term is for each.
It turns out that the first option gives a middle term of 2x.

Hence, your answer is (5x - 3)*(x + 1).

In general, the way to approach factoring is by looking at the coefficients of the first and last terms, and seeing how you can split them up. Then you kinda use trial and error to figure out where to go... there's no neat algorithm that'll take you straight to the correct solution, but with practice you'll learn to find it pretty quickly.

Hope that helps!

Answer 7

hahaha is this your home work?

1. 5x^2+2x-3
(5x-3)(x+1)

2. 6a^2 +5a+1
(3a+1)(2a+1)

3. 8b^2+2b-3
(4b+3)(2b-1)

4. 2a^2+19a-10
(2a-1)(a+10)

5. 6y^2-11y-10
(3y+2)(2y-5)

for explanations, i cant explain coz these are basic...need to figure it out for yourself...sorry

Answer 8

Hi,

I'm going to explain how I teach these, which may be different from the way your teacher explained them. Don't try to mix methods. It will usually mess things up.

1. 5x^2+2x-3

5x^2 + 2x - 3 Look for a GCF None this time. If there was one, factor it out. Then, temporarily start both parentheses with the first number and variable.

(5x.......)(5x..........) First sign goes in first parentheses.
(5x..+....)(5x..........) Product of signs goes in 2nd parentheses.
(5x.+....)(5x...-.....) <== minus is because pos x neg = negative

Now multiply your first and third numbers of the equation together. Ignore their signs - you've already done them.
5 x 3 = 15 So, out to the side list pairs of factors of 15.

15
------
1, 15
3, 5

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 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 to decide this.)

(5x.+....)(5x.-.....) Your signs are different, so you want to subtract factors to get 2. Those factors are 3 and 5. When you put the numbers in the parentheses, the bigger number is pushy - it always goes first. So your factors now are:
(5x.+.5)(5x.-.3)

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 5 but the second parentheses does not reduce.

(5x.+.5)(5x.-.3)
----------
......5 This reduces to your final factors of

(x.+.1)(5x.-.3)

2. 6a^2 +5a+1

6a^2 + 5a + 1 Look for a GCF None this time. If there was one, factor it out. Then, temporarily start both parentheses with the first number and variable.
(6a.......)(6a..........) First sign goes in first parentheses.
(6a..+....)(6a..........) Product of signs goes in 2nd parentheses.
(6a.+....)(6a..+.....) <== plus is because pos x pos = positive

Now multiply your first and third numbers of the equation together. Ignore their signs - you've already done them.
6 x 1 = 6 So, out to the side list pairs of factors of 6.

6
------
1, 6
2, 3

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 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 to decide this.)

(6a.+....)(6a.+.....) Your signs are the same, so you want to add factors to get 5. Those factors are 2 and 3. When you put the numbers in the parentheses, the bigger number is pushy - it always goes first. So your factors now are:
(6a.+..3)(6a.+.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 and the second parentheses is divisible by 2.

(6a.+..3)(6a.+.2)
----------..-----------
......3............2....... This reduces to your final factors of

(2a.+.1)(3a.+.1)


3. 8b^2+2b-3

8b^2 + 2b - 3 Look for a GCF None this time. If there was one, factor it out. Then, temporarily start both parentheses with the first number and variable.
(8b.......)(8b..........) First sign goes in first parentheses.
(8b..+....)(8b..........) Product of signs goes in 2nd parentheses.
(8b.+...)(8b...-....) <== minus is because pos x neg = negative

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

24
------
1, 24
2, 12
3, 8
4,6

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 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 to decide this.)

(8b.+...)(8b..-....) Your signs are different, so you want to subtract factors to get 2. Those factors are 4 and 6. When you put the numbers in the parentheses, the bigger number is pushy - it always goes first. So your factors now are:
(8b.+.6)(8b..-.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 2 and the second parentheses is divisible by 4.

(8b.+.6)(8b..-.4)
----------..----------
......2..........4 This reduces to your final factors to:

(4b.+.3)(2b..-.1)



4. 2a^2+19a-10

2a^2 + 19a - 10 Look for a GCF None this time. If there was one, factor it out. Then, temporarily start both parentheses with the first number and variable.
(2a.......)(2a..........) First sign goes in first parentheses.
(2a..+....)(2a..........) Product of signs goes in 2nd parentheses.
(2a.+...)(2a.-...) <== minus is because pos x neg = negative

Now multiply your first and third numbers of the equation together. Ignore their signs - you've already done them.
2 x 10 = 20 So, out to the side list pairs of factors of 20.

20
------
1, 20
2, 10
4,5

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 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 to decide this.)

(2a.+...)(2a.-...) Your signs are different, so you want to subtract factors to get 19. Those factors are 1 and 20. When you put the numbers in the parentheses, the bigger number is pushy - it always goes first. So your factors now are:
(2a.+.20)(2a.-.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 and the second parentheses does not reduce.

(2a.+.20)(2a.-.1)
------------
......2................. This reduces to your final factors to:

(a.+.10)(2a.-.1)


5. 6y^2-11y-10

6y^2 - 11y - 10 Look for a GCF None this time. If there was one, factor it out. Then, temporarily start both parentheses with the first number and variable.
(6y.......)(6y..........) First sign goes in first parentheses.
(6y..-....)(6y..........) Product of signs goes in 2nd parentheses.
(6y.-...)(6y.+...) <== plus is because neg x neg = positive

Now multiply your first and third numbers of the equation together. Ignore their signs - you've already done them.
6 x 10 = 60 So, out to the side list pairs of factors of 60.

60
------
1, 60
2, 30
3, 20
4,15
5,12
6,10

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 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 to decide this.)

(6y.-...)(6y.+...) Your signs are different, so you want to subtract factors to get 11. Those factors are 4 and 15. When you put the numbers in the parentheses, the bigger number is pushy - it always goes first. So your factors now are:
(6y.-.15)(6y.+.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 and the second parentheses is divisible by 2.

(6y.-.15)(6y.+.4)
-----------..---------
......3.............2... This reduces to your final factors to:

(2y.-.5)(3y.+.2)


I hope that helps and makes sense to you!! :-)