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Sagot :
Hmm… it would've been easier if you said a certain problem, because there can be a lot of ways to factor a binomial. But I'll do my best.
I'll use this binomial as an example: [tex](18 x^{10} + 54 x^{5})[/tex]
You can factor this by getting the Common Monomial Factor. It's kinda like GCF. This becomes [tex]18 x^{5} ( x^{2} + 3)[/tex] .
You can start with the coefficients (the numbers) first. The GCF of 18 and 54 is 18. So you put 18 outside of the parentheses. Then, you divide each number in the parentheses by 18. [tex]18 (1 + 3)[/tex]
Then you do same to the variables (the letters). To get the GCF of the variables, you get the variable that has the exponent that is smaller or less than the other(s), which is [tex] x^{5} [/tex] . Then you put that variable out of the parentheses beside 18. After that you divide each variable in the parentheses by [tex] x^{5} [/tex] . [tex] x^{5} ( x^{2} + 1)[/tex].
You put each term together then voila! you get [tex](18 x^{10} + 54 x^{5})[/tex] .
Some binomials are Special Products. The ones that I know are Difference of Two Squares, Difference of Two Cubes and Sum of Two Cubes. For me, the easiest is difference of two squares so I'll start with that first.
A difference of two squares looks like this: [tex]4 x^{2} - 9 x^{4} [/tex] . That will make
[tex](2 x - 3 x^{2})(2x + 3 x^{2} )[/tex] .
To do that you just get the square root of each term in the differences of two squares then you put them in a binomial with a minus (-) between them, then you put them in another binomial with a plus (+) in between them.
Okay I'm pretty sleepy already. I think another person will do the rest. Hope I helped! :)
I'll use this binomial as an example: [tex](18 x^{10} + 54 x^{5})[/tex]
You can factor this by getting the Common Monomial Factor. It's kinda like GCF. This becomes [tex]18 x^{5} ( x^{2} + 3)[/tex] .
You can start with the coefficients (the numbers) first. The GCF of 18 and 54 is 18. So you put 18 outside of the parentheses. Then, you divide each number in the parentheses by 18. [tex]18 (1 + 3)[/tex]
Then you do same to the variables (the letters). To get the GCF of the variables, you get the variable that has the exponent that is smaller or less than the other(s), which is [tex] x^{5} [/tex] . Then you put that variable out of the parentheses beside 18. After that you divide each variable in the parentheses by [tex] x^{5} [/tex] . [tex] x^{5} ( x^{2} + 1)[/tex].
You put each term together then voila! you get [tex](18 x^{10} + 54 x^{5})[/tex] .
Some binomials are Special Products. The ones that I know are Difference of Two Squares, Difference of Two Cubes and Sum of Two Cubes. For me, the easiest is difference of two squares so I'll start with that first.
A difference of two squares looks like this: [tex]4 x^{2} - 9 x^{4} [/tex] . That will make
[tex](2 x - 3 x^{2})(2x + 3 x^{2} )[/tex] .
To do that you just get the square root of each term in the differences of two squares then you put them in a binomial with a minus (-) between them, then you put them in another binomial with a plus (+) in between them.
Okay I'm pretty sleepy already. I think another person will do the rest. Hope I helped! :)
To make it short, I'll give you an example:
10x^2 + 5x
In this given you have to find the GCF of both in order to make it the shortest equation ever. The GCF is 5 right? However, you must also include a variable, since there are both x in the sides. So, the answer is 5x(2x + 1). Remember to follow the rule of sign!
Another given:
36x^4 - 9x^3
In this given, the GCF would be 9x^3, since if the answer is 9x^3(4x - 1), multiplying it makes the answer!
10x^2 + 5x
In this given you have to find the GCF of both in order to make it the shortest equation ever. The GCF is 5 right? However, you must also include a variable, since there are both x in the sides. So, the answer is 5x(2x + 1). Remember to follow the rule of sign!
Another given:
36x^4 - 9x^3
In this given, the GCF would be 9x^3, since if the answer is 9x^3(4x - 1), multiplying it makes the answer!
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