In order to factor x ^ 2 + 5x + 6 we must break up the “bx” term into two terms whose coefficients

In order to factor x ^ 2 + 5x + 6 we must break up the “bx” term into two terms whose coefficients…
add to 5 and multiply to 6
add to 6 and multiply to 5.
add to 1 and multiply to f.
add to 1 and multiply to 5.

The correct answer and explanation is:

The correct answer is: add to 5 and multiply to 6.

To factor the quadratic expression x2+5x+6x^2 + 5x + 6, we start by looking for two numbers that multiply to give the constant term (6) and add up to give the coefficient of the linear term (5).

1. The quadratic is in the standard form ax2+bx+cax^2 + bx + c, where:
   • a=1a = 1 (the coefficient of x2x^2),
   • b=5b = 5 (the coefficient of xx),
   • c=6c = 6 (the constant term).
2. We need two numbers that meet the following criteria:
   • Their product is c=6c = 6,
   • Their sum is b=5b = 5.
3. The possible pairs of numbers that multiply to 6 are:
   • 1 and 6 (since 1×6=61 \times 6 = 6),
   • 2 and 3 (since 2×3=62 \times 3 = 6).
4. Now, check which pair adds up to 5:
   • 1+6=71 + 6 = 7 (not 5),
   • 2+3=52 + 3 = 5 (this works).

So, the correct pair is 22 and 33. These numbers both multiply to 6 and add up to 5.

5. Now, break up the middle term 5x5x into 2x+3x2x + 3x and rewrite the quadratic as: x2+2x+3x+6.x^2 + 2x + 3x + 6.
6. Factor by grouping:
   • Group the first two terms: x2+2xx^2 + 2x,
   • Group the last two terms: 3x+63x + 6.
   Factor each group: x(x+2)+3(x+2).x(x + 2) + 3(x + 2).
7. Finally, factor out the common binomial factor (x+2)(x + 2): (x+2)(x+3).(x + 2)(x + 3).

Thus, the factorization of x2+5x+6x^2 + 5x + 6 is (x+2)(x+3)(x + 2)(x + 3).