Question

factoring ( x^2 + bx + c )
for 1-6, factor each trinomial if possible. if not possible, explain why.

1. ( x^2 + 11x + 10 )
2. ( x^2 - 12x + 27 )
3. ( x^2 - 9x + 5 )
4. ( x^2 + x - 20 )
5. ( x^2 + 6x - 2 )
6. ( x^2 - 7x - 44 )

Explanation:

Problem 1: \( x^2 + 11x + 10 \)

Step1: Find two numbers that multiply to 10 and add to 11.

The numbers are 1 and 10, since \( 1 \times 10 = 10 \) and \( 1 + 10 = 11 \).

Step2: Factor the trinomial.

Using the numbers from Step 1, we can write the trinomial as \( (x + 1)(x + 10) \).

Step1: Find two numbers that multiply to 27 and add to -12.

The numbers are -3 and -9, since \( (-3) \times (-9) = 27 \) and \( (-3) + (-9) = -12 \).

Step2: Factor the trinomial.

Using the numbers from Step 1, we can write the trinomial as \( (x - 3)(x - 9) \).

Step1: Check if we can find two numbers that multiply to 5 and add to -9.

The factors of 5 are 1 and 5, or -1 and -5.

• \( 1 + 5 = 6

eq -9 \)

• \( -1 + (-5) = -6

eq -9 \)
Since there are no two integers that multiply to 5 and add to -9, the trinomial cannot be factored using integer coefficients.

Answer:

\( (x + 1)(x + 10) \)

Problem 2: \( x^2 - 12x + 27 \)