Question

I am giving you challenge.
Who can solve this equation?
$\sf 2 {x}^{3} + 15 {x}^{2} + 37x + 30 = 0$
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Answer 1

Required Answer:-

Given:

• 2x³ + 15x² + 37x + 30 = 0

To find:

• The roots of the given equation.

Answer:

• The roots of the given equation are -2, -3 and -2.5

Solution:

We have,

➡ 2x³ + 15x² + 37x + 30 = 0

➡ 2x³ + 4x² + 11x² + 22x + 15x + 30 = 0

➡ 2x²(x + 2) + 11x(x + 2) + 15(x + 2) = 0

➡ (x + 2)(2x² + 11x + 15) = 0

➡ (x + 2)[2x² + 6x + 5x + 15] = 0

➡ (x + 2)[2x(x + 3) + 5(x + 3)] = 0

➡ (x + 2)(x + 3)(2x + 5) = 0

By zero product rule,

➡ Either x + 2 = 0 or x + 3 = 0 or 2x + 5 = 0

So,

➡ x + 2 = 0

➡ x = -2

Now,

➡ x + 3 = 0

➡ x = -3

Again,

➡ 2x + 5 = 0

➡ x = -2.5

Hence, the roots of the given cubic equation are -2, -3 and -2.5

Answer:

• The roots of the given cubic equation are -2, -3 and -2.5