Factor completely 6x^2-19x+14
To factor the given expression, we need to find two binomials whose product equals 6x^2 - 19x + 14.
To do this, we look for two binomials in the form (ax + b)(cx + d). Since 6x^2 is the leading term, we know that ac = 6.
Factors of 6:
1 * 6
2 * 3
Now, we need to find the values of b and d. Factors of 14:
1 * 14
2 * 7
We need to determine which combination of factors for ac and bd adds up to -19x, the coefficient of the middle term. Since the negative sign is involved, we will need a negative value.
Given that ac = 6:
1. ac pairs: (1,6) and (6,1)
- b and d could be the factors of 1 and 6, respectively: (1, 6), which would give (1x + 6)(6x + 1). However, this does not result in -19x when multiplied.
- b and d could be the factors of 6 and 1, respectively: (6, 1), which would give (6x + 1)(1x + 6). However, this also does not result in -19x when multiplied.
2. ac pairs: (2,3) and (3,2)
- b and d could be the factors of 2 and 3, respectively: (2, 3), which would give (2x + 3)(3x + 2). However, this does not result in -19x when multiplied.
- b and d could be the factors of 3 and 2, respectively: (3, 2), which would give (3x + 2)(2x + 3). Bingo! Now let's confirm if this multiplication results in -19x.
(3x + 2)(2x + 3):
6x^2 + 9x + 4x + 6
6x^2 + 13x + 6
The binomials (3x + 2)(2x + 3) factorize the quadratic expression 6x^2 - 19x + 14 completely.