How do you factor #6x^2+19x+3#?

Question

Eva Adamson · Accepted Answer

Use the new AC Method to factor. #f(x) = 6x^2 + 19x + 3 = (x - p)(x - q)#
Converted trinomial:# f'(x) = x^2 + 19x + 18 = (x - p')(x - q').#
Compose factor pairs of (a.c) = 18. Proceed (1, 18) = 19 = b. Then p' = 1 and q' = 18. We get: #p = (p')/a = 1/6 #and #q = (q')/a = 18 /6 = 3#.
Factored form:# f(x) = (x + 1/6)(x + 3) = (6x + 1)(x + 3)#
Check: Develop# f(x) = 6x^2 + 18x + x + 3 = 6x^2 + 19x + 3 #. Correct Reminder about the Rule of Signs.
a. When a and c have same signs, compose factor pairs of c, or (a.c), with all positive numbers.
Example: f(x) = x^2 + 51x + 98 = (x - p)(x - q). Compose factor pairs of (a.c = 98). a and are both positive. Proceed: (1, 98)(2, 49). This last sum is 51 = b. Then p = 2 and q = 49. b. When a and c have different signs, compose factor pairs of c, or (a.c), with all first numbers being negative.
Example. f(x) = x^2 - 11x - 42 = (x - p)(x - q). Compose factor pairs of (c = -42). Proceed: (-1, 42)(-2, 21)(-3, 14). This last sum is 11 = -b. Then the 2 real roots are -3 and 14.

Nathan Axel · Answer

undefined To factor the quadratic expression $6x^2 + 19x + 3$, you can use the factoring by grouping method or the quadratic formula. Let's use the factoring by grouping method:

1. Multiply the leading coefficient (6) by the constant term (3): $6 \times 3 = 18$.
2. Find two numbers that multiply to 18 and add to the coefficient of the middle term (19). The numbers are 1 and 18.
3. Rewrite the middle term (19x) using the two numbers found in the previous step: $19x = 1x + 18x$.
4. Group the terms: $6x^2 + 1x + 18x + 3$.
5. Factor by grouping: 
$(6x^2 + 1x) + (18x + 3)$.
$x(6x + 1) + 3(6x + 1)$.
6. Notice that $6x + 1$ is common to both terms. Factor it out: $(6x + 1)(x + 3)$.

So, the factored form of $6x^2 + 19x + 3$ is $(6x + 1)(x + 3)$.

Julia Adams · Answer

undefined To factor the expression $6x^2 + 19x + 3$, you can use the factoring method for trinomials.

The factored form of the expression is $(2x + 1)(3x + 3)$.