Question

Factor the polynomial, x2 + 5x + 6

completely (show work)

a. (x + 6)(x + 1)
b. (x + 3)(x + 2)
c. (x - 3)(x - 2)
d. (x-6)(x + 1)

Answer 1

Choice b.

`x^(2) + 5\, x + 6 = (x + 3)\, (x + 2)`.

Explanation:

The highest power of the variable
`x` in this polynomial is
`2`. In other words, this polynomial is quadratic.

It is thus possible to apply the quadratic formula to find the "roots" of this polynomial. (A root of a polynomial is a value of the variable that would set the polynomial to
`0`.)

After finding these roots, it would be possible to factorize this polynomial using the Factor Theorem.

Apply the quadratic formula to find the two roots that would set this quadratic polynomial to
`0`. The discriminant of this polynomial is
`(5^(2) - 4 * 1 * 6) = 1`.

`\begin{aligned}x_(1) &= (-5 + √(1))/(2* 1) \\ &= (-5 + 1)/(2) \\ &= -2\end{aligned}`.

Similarly:

`\begin{aligned}x_(2) &= (-5 - √(1))/(2* 1) \\ &= (-5 - 1)/(2) \\ &= -3\end{aligned}`.

By the Factor Theorem, if
`x = x_(0)` is a root of a polynomial, then
`(x - x_0)` would be a factor of that polynomial. Note the minus sign between
`x` and
`x_(0)`.

• The root
  `x = -2` corresponds to the factor
  `(x - (-2))`, which simplifies to
  `(x + 2)`.
• The root
  `x = -3` corresponds to the factor
  `(x - (-3))`, which simplifies to
  `(x + 3)`.

Verify that
`(x + 2)\, (x + 3)` indeed expands to the original polynomial:

`\begin{aligned}& (x + 2)\, (x + 3) \\ =\; & x^(2) + 2\, x + 3\, x + 6 \\ =\; & x^(2) + 5\, x + 6\end{aligned}`.