Is 3x^2+13x-10 factorable?? i dont think so but i need some back up... i am unsure?

Question

im sorry im just not seeing how -2 and 5 make a +13....

Answer 1

First set down factors of 3x²:-
(3x******) (x******)
Now require factors of 10 that combine with 3x and x to give 13x:-
(3x******2) (x******5)
Decide upon signs:-
(3x - 2) (x + 5)
Check
3x (x + 5) - 2 (x + 5)
3x² + 15x - 2x - 10
3x² + 13x - 10 (as required)

Answer 2

okay, you start off by factoring 3x^2. the way I set it up is this

(3x _ _) (x _ _)

the blanks obviously stand for the sign and number.


next you factor the -10. factors being -2 and 5.

so the pair now looks like this

(3x-2)(x+5)




"im sorry im just not seeing how -2 and 5 make a +13...."

the reason (3x-2)(x+5) equals 13x is because:


3x*5= 15x
-2*x= -2x

15x+ (-2x) = 13x.

hope that cleared some up. :]

Answer 3

To read what is written here, copy and paste the lines starting with \[ and ending in \] into the online form at http://www.tlhiv.org/cgi-bin/LaTeXpreviewer/index.cgi and press 'Preview'.

To find out, you need to simply solve for a quantity called the determinant. In any such second order equation:
\[
ax^2+bx+c
\]
\[
\Delta = b^2-4ac
\]

If
\[
\Delta \geq 0
\]
the expression is factorable. Otherwise it is not.

In your case, for example,

\[
\Delta = {13}^{2} + 4\times3\times{10} = 289 > 0
\]

So it is clearly factorable.

I can factor this therefore as:

\[
3{x}^{2}+13x-10 = 3\left( x-\frac{2}{3} \right) \left( x+5 \right)
\]

Answer 4

Actually, it is factorable. I can not explain it to you because I don't know if you are familiar with the method I use. However, I can tell you that the answer is:

(3x-2) (x+5)

Check it and you will see that I am correct. [:

Answer 5

Hi,

On every type of factoring problem ALWAYS look first for a GCF, a greatest common factor. There won't always be one, but if there is, you should divide it out first. A GCF will carry along through the problem and be the first thing in an answer with a GCF.

I' m going to explain factoring trinomials 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.

3x² + 13x - 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.
(3x.......)(3x..........) First sign goes in first parentheses.
(3x..+....)(3x..........) Product of signs goes in 2nd parentheses.
(3x..+....)(3x...-.....) <== neg is because pos x neg = negative

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.)

(3x..+....)(3x...-.....) 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:
(3x.+.15)(3x.-.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.

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

(x + 5)(3x - 2)<==THIS IS THE ANSWER TO YOUR PROBLEM.

Here is another problem so you can see the pattern of factoring!!

2x² +15x + 7 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.
(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!! :-)

Answer 6

3x^2+13x-10=0
multiply 10*3
3x^2x+15x-2x-10=0
3x(x+5)-2(x+5)=0
(3x-2) (X+5)=0

Answer 7

3x^2+13x-10
no not factorisable

Answer 8

no its factorizable
its factors are
3x+5 * x-2
c its very simpe

Answer 9

(3x-2)(x+5)

Answer 10

4x-10