$4q^2+2q=3q^2-4q+16\qquad\text{subtract}\ 3q^2\ \text{from both sides}\\\\q^2+2q=-4q+16\qquad\text{add}\ 4q\ \text{to both sides}\\\\q^2+6q=16\qquad\text{subtract 16 from both sides}\\\\q^2+6q-16=0\to\boxed{\text{standard form}}\\\\q^2+8q-2q-16=0\\\\q(q+8)-2(q+8)=0\\\\(q+8)(q-2)=0\iff q+8=0\ \vee\ q-2=0\\\\q+8=0\qquad\text{subtract 8 from both sides}\\\\\boxed{q=-8}\\\\q-2=0\qquad\text{add 2 to both sides}\\\\\boxed{q=2}\\\\Answer:\ \boxed{q=-8\ or\ q=2}$

3 0

4 years ago

125x^3-27 factor polynomials

vampirchik [111]

Answer:

The answer is (5x - 3) (25x² + 15x + 9)

Step-by-step explanation:

125x³ - 27

5 × 5 × 5 × x × x × x - 3 × 3 × 3

Rewrite 125x³ as (5x)³.

(5x)³ - 27

Rewrite 27 as (3)³.

(5x)³ - (3)³

Here, Both terms are perfect cubes, factor using the difference of cube formula,

a³ - b³ = (a - b) (a² + ab + b²)

where,

a = 5x

b = 3

Now,

(5x - 3) [(5x)² + (5x)(3) + (3)²]

Now, Simplify

(5x - 3) (25x² + 15x + 9)

Thus, The answer is (5x - 3) (25x² + 15x + 9)

-TheUnknownScientist 72

4 0

3 years ago