directions are: Factor each of the following polynomials completely.
14x^2-20x+6
I am at this almost last point but i don't know what to do after or if this is the answer
7x^-10x+3=2(7x-3)(x-1)
your final answer is correct
I am going w/ Anonymous.....
It's correct!
To factor the polynomial 14x^2 - 20x + 6 completely, you can follow these steps:
Step 1: Look for any common factors:
In this case, there are no common factors among the coefficients (14, -20, and 6).
Step 2: Check if it can be factored using the quadratic formula:
To determine if the polynomial can be factored using the quadratic formula, calculate the discriminant (b^2 - 4ac). For the polynomial 14x^2 - 20x + 6, a = 14, b = -20, and c = 6.
Discriminant = (-20)^2 - 4(14)(6) = 400 - 336 = 64.
Since the discriminant is positive (64 > 0), the polynomial can be factored.
Step 3: Use the quadratic formula to find the roots:
The roots of the polynomial can be found using the quadratic formula, which states that for a quadratic equation ax^2 + bx + c = 0, the roots can be calculated as:
x = (-b ± √(b^2 - 4ac)) / 2a.
For the polynomial 14x^2 - 20x + 6, a = 14, b = -20, and c = 6. Plugging these values into the quadratic formula, we get:
x = (20 ± √(20^2 - 4(14)(6))) / (2(14)).
Simplifying further:
x = (20 ± √(400 - 336)) / 28
x = (20 ± √64) / 28
x = (20 ± 8) / 28
Hence, the roots of the polynomial are x = (20 + 8) / 28 =28/28= 1 and x = (20 - 8) / 28 = 12 / 28 = 3/7.
Step 4: Write the factored form using the roots:
Since the roots of the polynomial are 1 and 3/7, the factored form can be written as:
14x^2 - 20x + 6 = 14(x - 1)(x - 3/7).
Therefore, the fully factored form of the polynomial 14x^2 - 20x + 6 is 14(x - 1)(x - 3/7).