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solve for all values of x by factoring $x^2 - 32 = -6x - 5$ assess the …

Question

solve for all values of x by factoring
$x^2 - 32 = -6x - 5$
assess the validity of your solutions.

Explanation:

Step1: Rearrange to standard quadratic form

$x^2 + 6x - 32 + 5 = 0$
$x^2 + 6x - 27 = 0$

Step2: Factor the quadratic expression

Find two numbers that multiply to $-27$ and add to $6$: $9$ and $-3$.
$(x + 9)(x - 3) = 0$

Step3: Solve for $x$ using zero product rule

Set each factor equal to 0:
$x + 9 = 0$ or $x - 3 = 0$
$x = -9$ or $x = 3$

Step4: Verify solutions (substitute back)

For $x=-9$:
Left side: $(-9)^2 - 32 = 81 - 32 = 49$
Right side: $-6(-9) - 5 = 54 - 5 = 49$
For $x=3$:
Left side: $3^2 - 32 = 9 - 32 = -23$
Right side: $-6(3) - 5 = -18 - 5 = -23$
Both satisfy the equation.

Answer:

$x = -9$ and $x = 3$