## Factoring Trinomials of the Form \(x^2 - bx + c\)

Factoring trinomials, which are polynomials with three terms, can initially seem challenging. However, with practice, it becomes routine. If a trinomial factors, it will factor into the product of two binomials.

### Steps to Factor Trinomials:

**Factor the first term:**\(x^2 = x \cdot x\).**Factor the last term:**Choose factors that multiply to obtain the last term and add to obtain the middle term.**Determine the signs:**by adding or subtracting the product of the inner and outer terms.**Check by multiplying.**

### Example:

Factor the trinomial \(x^2 - 7x + 10\).

- Factor the first term: \(x^2 = x \cdot x\).
- Factor the last term: \(10 = 5 \cdot 2\) and \(5 + 2 = 7\), which is our middle term.
- Determine the signs: Both signs are negative because \(-5 - 2 = -7\).
- Check by multiplying: \((x - 5)(x - 2) = x^2 - 7x + 10\).

So, \(x^2 - 7x + 10 = (x - 5)(x - 2)\).

### Exercises:

Factor the following trinomials:

- \(x^2 - 5x + 6\)
- \(x^2 - 9x + 20\)
- \(x^2 - 7x + 10\)
- \(x^2 - 11x + 30\)
- \(x^2 - 8x + 15\)
- \(x^2 - 10x + 21\)
- \(x^2 - 6x + 8\)
- \(x^2 - 12x + 35\)
- \(x^2 - 15x + 56\)
- \(x^2 - 13x + 40\)

### Solutions:

- \(x^2 - 5x + 6 = (x - 3)(x - 2)\)
- \(x^2 - 9x + 20 = (x - 5)(x - 4)\)
- \(x^2 - 7x + 10 = (x - 5)(x - 2)\)
- \(x^2 - 11x + 30 = (x - 6)(x - 5)\)
- \(x^2 - 8x + 15 = (x - 5)(x - 3)\)
- \(x^2 - 10x + 21 = (x - 7)(x - 3)\)
- \(x^2 - 6x + 8 = (x - 4)(x - 2)\)
- \(x^2 - 12x + 35 = (x - 7)(x - 5)\)
- \(x^2 - 15x + 56 = (x - 8)(x - 7)\)
- \(x^2 - 13x + 40 = (x - 8)(x - 5)\)

Remember to always check your results by multiplying the factors. If the product matches the original trinomial, then the factoring is correct.