The inequality x2 + 12x + 35 ≥ 0 has two critical points and three possible intervals for solutions. choose each set of possible test points for the three intervals. –8, –6, –4 –10, –6, 0 –6, 0, 6 –6, 0, 10

-8,-6,-4

-10,-6,0

these are the possible test points for your three intervals

The given inequality is, x² + 12 x+35≥0

→ x²+7 x +5 x+5×7≥0

→ x (x+7) +5(x+7)≥0

→ (x+5)(x+7)≥0

→ x+5=0  ∧ x+7=0 gives x= -7 ∧ x= -5

Now drawing the number line and marking point ,-5 and -7 on it.

Now the three intervals are (-∞ , -7], [-7,-5] and [-5,∞).

The solution set of inequality (x+5)(x+7)≥0 is  (-∞ , -7] and [-5,∞).

The set of possible test points for

⇒ (-∞ , -7] → -8,  -10

⇒ [-7,-5] → -6

⇒ [-5,∞) → -4, 0

Option (1) -8,-6,-4 and Option (2) -10,-6,0 satisfies the given condition.