is 3n^2 - 11n + 6 factorable?!
all i know is you set it up 3(n^2-11n=6)
sorry not =6 but +6 rather
im sorry but taysha is wrong in her explanation...
it is factorable
3n^2 - 11n + 6 = (3n - 2)(n-3)
To determine if the expression 3n^2 - 11n + 6 is factorable, we can first check if it can be factored using the quadratic formula. The quadratic formula states that for a quadratic equation of the form ax^2 + bx + c = 0, the solutions for x can be found using the formula:
x = (-b ± √(b^2 - 4ac)) / 2a.
In the equation 3n^2 - 11n + 6, a = 3, b = -11, and c = 6. Let's substitute these values into the quadratic formula to find the solutions for n:
n = (-(-11) ± √((-11)^2 - 4(3)(6))) / (2(3))
= (11 ± √(121 - 72)) / 6
= (11 ± √49) / 6
= (11 ± 7) / 6.
Simplifying further, we have two possible solutions for n:
n = (11 + 7) / 6 = 18 / 6 = 3,
n = (11 - 7) / 6 = 4 / 6 = 2/3.
Since the equation has two distinct solutions, it indicates that it can be factored.
To find the factors of the quadratic expression 3n^2 - 11n + 6, we can use the factored form, which can be determined by setting the expression equal to zero:
3n^2 - 11n + 6 = 0.
Next, find two numbers, let's call them a and b, that multiply together to give ac (in this case, ac = 3 * 6 = 18) and add up to give b (in this case, b = -11).
In this case, the numbers that satisfy these conditions are 9 and 2.
Now, we rewrite the original equation as follows, replacing the middle term -11n with the sum of two terms using 9n and 2n:
3n^2 - 9n - 2n + 6 = 0.
Next, factor by grouping:
(3n^2 - 9n) + (-2n + 6) = 0.
Taking out the greatest common factor from the first two terms and the last two terms:
3n(n - 3) - 2(n - 3) = 0.
We can see that we now have the term "n - 3" appearing in both groups. Factoring out the common factor, we get:
(n - 3)(3n - 2) = 0.
So, the factored form of the quadratic equation 3n^2 - 11n + 6 is (n - 3)(3n - 2).