I wrote down some things I don't get about factoring quadratic equations...
1. When you find a common factor (when factoring) to you take the number and the letter out of the bracket, or just the number?
2. If I get a question like n^2 -1 or a^2 - 36, etc is it already factored?
3. I CANNOT FOR THE LIFE OF ME FIGURE OUT HOW TO FACTOR:
Y = 3x^2 + 14x + 8
Thank you, I think I might almost get factoring!
1. When you find a common factor (when factoring) to you take the number and the letter out of the bracket, or just the number?
2. If I get a question like n^2 -1 or a^2 - 36, etc is it already factored?
3. I CANNOT FOR THE LIFE OF ME FIGURE OUT HOW TO FACTOR:
Y = 3x^2 + 14x + 8
Thank you, I think I might almost get factoring!
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1. If you have both a common letter/variable and a number you take both out. Otherwise, only the one that is common (i.e. 9a^2 -6a) = 3a(3a - 2)
2. n^2 - 1 = (n+1)(n-1) , a^2 - 36 = (a+6)(a-6)
3. 3x^2 + 14x + 8 = (3x + 2)(x + 4), if you can find the values of x that satisfy the equation for it to be 0 then is a lot easier. Additionally, you can do the following:
3x^2 + 14x + 8 = 3x^2 + 12x + 2x + 8 = 3x(x + 4) + 2(x + 4) = (x + 4)(3x + 2) but is too difficult to notice something like that.
2. n^2 - 1 = (n+1)(n-1) , a^2 - 36 = (a+6)(a-6)
3. 3x^2 + 14x + 8 = (3x + 2)(x + 4), if you can find the values of x that satisfy the equation for it to be 0 then is a lot easier. Additionally, you can do the following:
3x^2 + 14x + 8 = 3x^2 + 12x + 2x + 8 = 3x(x + 4) + 2(x + 4) = (x + 4)(3x + 2) but is too difficult to notice something like that.
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1. i dont understand the question...
2. if an equation is factored, it will be inside brackets. eg> (x-1)(x+4) <
3. multiply the 'a' and 'c' value together. it equals 24.
then find two numbers that multiply to 24 and add to the 'b' value.
the two numbers are 12 and 2
take these numbers and put them into the equation like this y = 3x^2 + 12x + 2x + 8.
then take out the common values in the first set. the first set is 3x^2 + 12x. so the common values are 3 and an 'x'. so it looks like this 3x(x+4).
do the common values to the next set.. 2x + 8. it will look like this 2(x+4)
it should look like this 3x(x+4) + 2(x+4)
the final answer is (3x+2)(x+4)
2. if an equation is factored, it will be inside brackets. eg> (x-1)(x+4) <
then find two numbers that multiply to 24 and add to the 'b' value.
the two numbers are 12 and 2
take these numbers and put them into the equation like this y = 3x^2 + 12x + 2x + 8.
then take out the common values in the first set. the first set is 3x^2 + 12x. so the common values are 3 and an 'x'. so it looks like this 3x(x+4).
do the common values to the next set.. 2x + 8. it will look like this 2(x+4)
it should look like this 3x(x+4) + 2(x+4)
the final answer is (3x+2)(x+4)
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When numbers don't have a common factor but are both square numbers (your example of n^2-1) you can see that it is a conjugate. Find the squareroot of both numbers (n and 1) and put them into brackets with a plus and a minus sign (n-1)(n+1)
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1) That's a pretty vague question. If you mean something like (27x^2+6x), that would be 3x(9x+2)
2) Hell no. n^2-1 is (n-1)(n+1). Go study your formulas.
3) Epic facepalm. (3x+2)(x+4).
2) Hell no. n^2-1 is (n-1)(n+1). Go study your formulas.
3) Epic facepalm. (3x+2)(x+4).
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Factoring is hard. The only test I got a 50 and not 90 on for gade 10 math was factoring. My advise study because even though we supply you the answers there is no doubt the book can help you more.