# For what values of x is the product (x+4)(x+ 6) positive?

Dec 16, 2016

$\left(x + 6\right) \left(x + 4\right) > 0 \text{ if " x> -4 " or } x < - 6$.

#### Explanation:

If we are excluding the product being 0 then we can say that $\left(x + 4\right) \left(x + 6\right) > 0$

Method 1
$x + 4 = 0 , x = - 4$
$x + 6 = 0 , x = - 6$

Through trial and error we can deduce that,
$\left(x + 6\right) \left(x + 4\right) > 0 \text{ if " x> -4 " or } x < - 6$.

Method 2
=(x(x) + x(6) + 4(x) + 4(6) > 0
${x}^{2} + 10 x + 24 > 0$

Now we can identify our abc values and solve for x using the quadratic formula.

$a = 1 , b = 10 , c = 24$
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
${x}_{1} = \frac{- 10 + \sqrt{{10}^{2} - 4 \left(1\right) \left(24\right)}}{2 \left(1\right)} , {x}_{2} = \frac{- 10 - \sqrt{{10}^{2} - 4 \left(1\right) \left(24\right)}}{2 \left(1\right)}$

${x}_{1} = \frac{- 10 + \sqrt{100 - 96}}{2} , {x}_{2} = \frac{- 10 - \sqrt{100 - 96}}{2}$

${x}_{1} = \frac{- 10 + \sqrt{4}}{2} , {x}_{2} = \frac{- 10 - \sqrt{4}}{2}$

${x}_{1} = \frac{- 10 + 2}{2} , {x}_{2} = \frac{- 10 - 2}{2}$

${x}_{1} = \frac{- 8}{2} , {x}_{2} = \frac{- 12}{2}$

${x}_{1} = - 4 , {x}_{2} = - 6$

Through trial and error we can deduce that,
$\left(x + 6\right) \left(x + 4\right) > 0 \text{ if " x> -4 " or } x < - 6$.

Dec 16, 2016

$x < - 6$ or $x > - 4$

#### Explanation:

$\left(x + 4\right) \left(x + 6\right)$ will be positive

(A) if both $\left(x + 4\right)$ and $\left(x + 6\right)$ are positive i.e. $x + 4 > 0$ and $x + 6 > 0$ i.e. $x > - 4$ and $x > - 6$. This is possible only if $x > - 4$.

or

(B) if both $\left(x + 4\right)$ and $\left(x + 6\right)$ are negative i.e. $x + 4 < 0$ and $x + 6 < 0$ i.e. $x < - 4$ and $x < - 6$. This is possible only if $x < - 6$.

Dec 16, 2016

x is outside $\left[- 6 , - 4\right]$.
The 1-D shaded x-axis illustrates this solution. The gap is out of bounds.

#### Explanation:

graph{(x+4)(x+6) > 0x^2 [-10, 10, -1, 1]}

$\left(x + 4\right) \left(x + 6\right) > 0$. So, the factors have the same sign.

And so, $x \succ 4 \mathmr{and} x > - 6 \to x > - 4$

and

x < -4 and x < -6 to x < -6#

Thus, $x < - 6 \mathmr{and} x > - - 4$