How do I factor 3x2+4x=1 and #x2-2x-8
thats the same as 6+4x=1
take out a 2 : 2(3+2x)=1
divide each side by 2: 3+2x=(1/2)
take 3 from each side: 2x= (-5/2)
then divide each side by 2 to get x:
x= -5/4
I have an impression that the character x stands for the variable, and not a multiplication sign. Also, the equal sign is meant to be a + sign, and the # is meant to be a three (just a guess).
The question should then be:
"How do I factor 3x2+4x+1 and 3x2-2x-8?"
1.
3x²+4x+1
From 3x² look for a pattern, since the only factors of three are 1 and 3.
(3x+a)(x+b)
Since a*b=1, a=1 and b=1 or a=-1 and b=-1.
Since the second term is +4x, we conclude that a=1 and b=1.
Try
(3x+1)(x+1)
2.
3x2-2x-8
Proceed as in #1,
(3x+a)(x+b)
Now, since the constant term -8 has multiple factors (1,2,4 and 8), we have to make more trial combinations that make a+3b=-2 (coefficient of -2x) and a*b=-8 (constant term).
Make a table of a,b, a+3b, a*b and calculate, using possible values of a and b as ±(1,2,4,8). Notice that since a+3b=-2, at least one of a or b has to be negative, also b=-8/a
a b a+3b a*b
1 -8 -23 -8
2 -4 -10 -8
4 -2 -2 -8 (we found the combination).
So the factors are:
(3x+4)(x-2)
check:
(3x+4)(x-2)=3x²-2x-8 OK.
To factor the quadratic equations 3x^2 + 4x = 1 and x^2 - 2x - 8, we can follow these steps:
1. 3x^2 + 4x = 1:
- First, move all terms to one side of the equation to make it equal to zero: 3x^2 + 4x - 1 = 0.
- To factor this quadratic equation, we need to find two integers whose product is the product of the leading coefficient (3) and the constant term (-1). In this case, the product is -3.
- Additionally, we need to find two integers whose sum is equal to the coefficient of the middle term (4). In this case, those two numbers are 3 and 1.
- Rewrite the middle term (4x) as the sum of the two numbers found in the previous step: 3x^2 + 3x + x - 1 = 0.
- Group the terms and factor by grouping: (3x^2 + 3x) + (x - 1) = 0.
- Factor out the greatest common factor from each group: 3x(x + 1) + (x - 1) = 0.
- Factor out a common term from both groups: (3x + 1)(x + 1) = 0.
- Set each factor equal to zero to solve for x: 3x + 1 = 0 or x + 1 = 0.
- Solve for x: x = -1 or x = -1/3.
2. x^2 - 2x - 8:
- Again, move all terms to one side to make it equal to zero: x^2 - 2x - 8 = 0.
- To factor this quadratic equation, we need to find two integers whose product is the product of the leading coefficient (1) and the constant term (-8). In this case, the product is -8.
- Additionally, we need to find two integers whose sum is equal to the coefficient of the middle term (-2). In this case, those two numbers are -4 and 2.
- Rewrite the middle term (-2x) as -4x + 2x: x^2 - 4x + 2x - 8 = 0.
- Group the terms and factor by grouping: (x^2 - 4x) + (2x - 8) = 0.
- Factor out the greatest common factor from each group: x(x - 4) + 2(x - 4) = 0.
- Factor out a common term from both groups: (x + 2)(x - 4) = 0.
- Set each factor equal to zero to solve for x: x + 2 = 0 or x - 4 = 0.
- Solve for x: x = -2 or x = 4.
So, the factored form of the quadratic equations is:
1. 3x^2 + 4x = 1: (3x + 1)(x + 1) = 0
2. x^2 - 2x - 8: (x + 2)(x - 4) = 0