how to factor math problems?

Question

c^2+17cd+70d^2

i have trouble explaing and working it out could someone show me the steps to this
and is there an easier way to solve it

Answer 1

Hi,

First , make sure your terms are in the correct order so one of the variables has its exponents in decreasing order. Your variable "c" is in descending order of exponents, going 2 - 1 - 0, so you're fine.
Next, does your problem have a GCF? If so, factor it out now. This time it doesn't, but you should always check for one. If there was one, you would carry it along and it would be the first thing in your factored answer.
Now make your 2 parentheses for your binomial factors. Take your first term and put its coefficient and the variable to half of its exponent in both parentheses as the first term. So you would just have ( c )(c ) The one in front is understood as the coefficient. Now the first sign in the problem goes in the first parentheses. The product of the 2 signs in the problem goes in the second parentheses. SInce a positive times a positive is still positive, both signs are plus this time. (c + )(c + )
Next multiply the numbers from your first and third terms together and write this number off to the side. You want to list out pairs of factors of this number. You will use one set of these in the back of the 2 factors. In this problem 1 x 70 = 70, so you want to look at pairs of factors of 70. They could be 1 x 70, 2 x 35, 5 x 14, or 7 x 10.
You select the factors you need by this rule. If the signs you put in the parentheses are the same (either BOTH positive or BOTH negative) then you want the pair of factors that ADD up to equal the center number of your problem. If the signs you put in the parentheses are different (one positive and one negative) then you want the pair of factors that SUBTRACT to equal the center number of your problem. Since your signs are the same, you want the pair of factors that ADD to 17, which is 7 x 10. The bigger number is pushy and always goes first, so your factors are (c + 10)(c + 7).
Now just like the c^2 at the front of the problem split into c and c at the front of your factors, d^2 spilts into d and d in the back of your factors, so they are (c + 10d)(c + 7d).
If there had been coefficients in front of the first term other than 1, you would have to reduce the factors to be done. This doesn't happen here, so you're done.

Suppose now you wanted to factor 8x^2 -10x - 3.
The terms are in the right order because exponents go 2 - 1 - 0 on x. Next, there is no GCF to factor out. Now make your 2 parentheses for your binomial factors. Take your first term and put its coefficient and the variable to half of its exponent in both parentheses as the first term. So you would just have ( 8x )(8x ) This isn't correct for the final factors, but it fixes itself in a minute! Now the first sign in the problem goes in the first parentheses. The product of the 2 signs in the problem goes in the second parentheses. SInce a negative times a negative is still positive, both signs are negative this time. (8x - )(8x - )
Next multiply the numbers from your first and third terms together and write this number off to the side. You want to list out pairs of factors of this number. Don't worry about their signs - you've already done signs. You will use one set of these in the back of the 2 factors. In this problem 8 x 3 = 24, so you want to look at pairs of factors of 24. They could be 1 x 24, 2 x 12, 3 x 8, or 4 x 6.
You select the factors you need by this rule. If the signs you put in the parentheses are the same (either BOTH positive or BOTH negative) then you want the pair of factors that ADD up to equal the center number of your problem. If the signs you put in the parentheses are different (one positive and one negative) then you want the pair of factors that SUBTRACT to equal the center number of your problem. Since your signs are the same, you want the pair of factors that ADD to 10, which is 4 x 6. The bigger number is pushy and always goes first, so your factors are (8x - 6)(8x - 4).
There is no second letter to put in this time. Since there are coefficients in front of the first term other than 1, you have to reduce the factors by the largest possible number in each factor to be done. Since 8 and 6 in the first factor are both divisible by 2, divide them by 2. This factor becomes (4x - 3). Divide the second factor by the largest possible number, 4. This factor becomes (2x - 1)
So your reduced correct factors are (4x - 3)(2x - 1).

Suppose now you wanted to factor 3x^2 -15xy + 18y^2.
The terms are in the right order because exponents go 2 - 1 - 0 on x. Next, there is a GCF of 3 to factor out. So your problem now is 3(x^2 -5xy + 6y^2). Now make your 2 parentheses for your binomial factors. Take your first term and put its coefficient and the variable to half of its exponent in both parentheses as the first term. So you would just have 3( x )( x )
Now the first sign in the problem goes in the first parentheses. The product of the 2 signs in the problem goes in the second parentheses. SInce a negative times a positive is a negative, both signs are negative this time. 3(x - )(x - )
Next multiply the numbers from your first and third terms after the GCF together and write this number off to the side. You want to list out pairs of factors of this number. Don't worry about their signs - you've already done signs. You will use one set of these in the back of the 2 factors. In this problem 1 x 6 = 6, so you want to look at pairs of factors of 6. They could be 1 x 6, or 2 x 3. Notice one pair subtracts to 5 while the other pair adds to 5, so you have to know the rule.
You select the factors you need by this rule. If the signs you put in the parentheses are the same (either BOTH positive or BOTH negative) then you want the pair of factors that ADD up to equal the center number of your problem. If the signs you put in the parentheses are different (one positive and one negative) then you want the pair of factors that SUBTRACT to equal the center number of your problem. Since your signs are the same, you want the pair of factors that ADD to 5, which is 2 x 3. The bigger number is pushy and always goes first, so your factors are 3(x - 3)(x - 2). Now the y^2 splits into y and y at the back to give 3(x - 3y)(x - 2y) Since there are no coefficients in front of the first term other than 1, you are done!
So your correct factors are 3(x - 3)(x - 2).

I know that was extra, but I hope it helps.

Answer 2

First, you need to factor the first problem (c^2) and it can only comes out as (c+??) and (c+??). Then the last problem (70d^2) which comes out multiple answers but you can know the right answer by calculating it. It goes like this, the factor of 70 is 1 and 70, 2 and 35, 5 and 14, 7 and 10, the middle (problem) must be the summary of the 2 factors. So the correct one is 10 and 7 because 10 plus 7 is 17 (which is the factor of the second problem) . So you'll have (c+10d) (c+7d) oh! and don't forget to put the second... (what was it called?) letters (I forgot the name for it)
I'm sorry if that doesn't make sense, English is not my first language and I'm just an 9th grader...

Answer 3

c^2 + 17cd + 70d^2
= (c + 10d)(c + 7d)
notice that 10 and 7 are factors of 70 and the sum of 10 and 7 is 17
so think of factors that add up to the middle
once you get the hang of this, you'll be factoring in no time.
good luck.

Answer 4

a(x+u)(x+v) = ax^2 + a(u+v)x + auv where u,v are -roots
set questions often have integer roots

sum roots = 17d
product of roots = 70d^2 = 2.5.7.d.d
2.5+7 = 17

(c + 10d)(c + 7d)

Answer 5

Look at the coeffecients 17 and 70. You're looking for

x+y=17
xy=70

Solve to get 7 and 10

(c+7d)(c+10d)

Answer 6

c^2 + 17cd + 70d^2
=c^2 + 10cd + 7cd + 70d^2
=c(c+10d) + 7d(c+10d)
=(c+10d)(c+7d)
= She + 10d and She + 7d is answer

Answer 7

c^2+10cd+7cd+70d^2
c(c+10d)+7d(c+10d)
(c+10d)(c+7d)