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Does the sight of a number or expression accompanied by the instructions, "Factor completely," strike fear into your heart? Wish you paid attention in algebra? Well, this instructable will teach you how to factor any number, or eligible expression such as Ax^2 + Bx + C.
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Signing UpStep 1Factoring numbers
First off, what is a factor?
"Natural number factors" are the complete set of whole numbers, where if you multiply one number in the set by another in the set, you get the number that you're factoring.
For example, the number 5 has two factors: 1, and 5. The number 6 has four factors: 1, 2, 3, and 6.
"Integer factors" include negative numbers.
The number 5 in this case would have four factors: -5, -1, 1, and 5. 6 would have eight factors: -6, -3, -2, -1, 1, 2, 3, and 6.
(Natural numbers are numbers without fractions, starting from 1, 2, 3, 4, 5... all the way up to infinity. Integers are natural numbers, as well as their negative counterparts and 0, or ...-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5...)
Factoring numbers with the natural number set is simple. Every number has at least two factors. To find other factors, start dividing the number starting from two and working your way up until you reach that number divided by 2. Any quotient that does not have a remainder means that both the divisor and the quotient are factors of that number.
Say you need to factor the number 9. You can't divide by two evenly, so we skip it. (Note the solution, 4.5, so you know when to stop later on.) 9 is divisible by 3, so add 3 to your list of factors. Work your way up until you divide by 5 (9 divided by 2, rounded up). You'll end up with 1, 3, and 9 as a list of factors.
When factoring numbers in the integer set, you can just add the negative equivalent of your solutions from natural number factoring in. So 9 would have factors of -9, -3, -1, 1, 3, and 9.
Factoring negative numbers can only be done with integer factoring. The solution is the same one you get factoring the positive version of the number. -9 has factors of -9, -3, -1, 1, 3, and 9.
Zero is the only integer that has an infinite amount of factors, and is the only one that has zero as a factor.
"Natural number factors" are the complete set of whole numbers, where if you multiply one number in the set by another in the set, you get the number that you're factoring.
For example, the number 5 has two factors: 1, and 5. The number 6 has four factors: 1, 2, 3, and 6.
"Integer factors" include negative numbers.
The number 5 in this case would have four factors: -5, -1, 1, and 5. 6 would have eight factors: -6, -3, -2, -1, 1, 2, 3, and 6.
(Natural numbers are numbers without fractions, starting from 1, 2, 3, 4, 5... all the way up to infinity. Integers are natural numbers, as well as their negative counterparts and 0, or ...-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5...)
Factoring numbers with the natural number set is simple. Every number has at least two factors. To find other factors, start dividing the number starting from two and working your way up until you reach that number divided by 2. Any quotient that does not have a remainder means that both the divisor and the quotient are factors of that number.
Say you need to factor the number 9. You can't divide by two evenly, so we skip it. (Note the solution, 4.5, so you know when to stop later on.) 9 is divisible by 3, so add 3 to your list of factors. Work your way up until you divide by 5 (9 divided by 2, rounded up). You'll end up with 1, 3, and 9 as a list of factors.
When factoring numbers in the integer set, you can just add the negative equivalent of your solutions from natural number factoring in. So 9 would have factors of -9, -3, -1, 1, 3, and 9.
Factoring negative numbers can only be done with integer factoring. The solution is the same one you get factoring the positive version of the number. -9 has factors of -9, -3, -1, 1, 3, and 9.
Zero is the only integer that has an infinite amount of factors, and is the only one that has zero as a factor.
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3(cube root) and then the radical sign and then 24n^2 x 3(cube root) and then the radical sign and then 36n^2..............
sorry i don't know how to make it look like the actual problem! How would I factor and solve that? Because obviously i am too dumb to figure it out on my own. online school is hard for me.
24n^2 = 36n^2
And then move the 24n^2 to the other side of the equation:
36n^2 - 24n^2
Factor out the n^2:
n^2(36 - 24) = 0
Solve.
n = 0
If you meant that the two terms were added together, that's unfactorable.
Know how to "factor" Binomials:
Problem: -3x^2 + 7x
Answer: 0, 7/3
I thought the obvious answer is pulling out the x for the original binominal which would result in x (-3x+7)
Although I haven't seen this before, I could take each of the factors and have them equal 0 to come up with the answer. Am I missing something?
Example: x = 0 and -3x + 7 = 0
Solve for x, x = 0 and x = 7/3
x^2(2+5x^5)(1+3x)
Answer has 3 factors (^^)
2x2 is strange looking, and could b thought of as 2 times x times 2
also, this gets more complicated when u have variable powers, as x to the power of y, which is different when written as xy.
ex:
x to the power of 4
wrong:
x4
right:
x^4
Hope this explains it, if not, reply to me
Square root both of the terms in the function. You'll end up with 2v and 9t.
Then, plug it in to the set pattern, (x + y)(x - y).
(2v + 9t)(2v - 9t)
(Imagine the tune of "Pop! Goes the Weasel".)
The opposite of lower case "B",
Plus or minus the square root,
Of "B" squared minus 4(ac),
All over 2a.
Fin.
It's irritating, but you'll remember the formula.
Pronounce...
4(ac) = four "A" "C"
2a = two "A"
x equals minus b,
plus or minus root,
b squared minus four a c,
all over two a.
:)
I mean 9.x.x times 9.x.x = 81.x.x.x.x
( sorry, can't type the power superscripts )
814 is 43,046,721
ax4+bx3+cx2+dx+e
synthetic division
normal division those would be reallly helpful.