Factorize 2x

For factorizing this cubic expression, no identity is useful.

Thus, we need to use factorization by regrouping terms.

We can write, –x² = x² – 2x² and –13x = –x – 12x

⇒ 2x³ – x² – 13x – 6 = 2x³ + x² – 2x² – x – 12x – 6

We have, –2x² = (–x) × 2x and –12x = (–6) × 2x

⇒ 2x³ – x² – 13x – 6 = 2x³ + x² + (–x) × 2x – x + (–6) × 2x – 6

Observe that x² is common for the first two terms, –x is common for the next two terms and –6 is common for the last two terms.

⇒ 2x³ – x² – 13x – 6 = x²(2x + 1) + (–x)(2x + 1) + (–6)(2x + 1)

Now, (2x + 1) is the common term.

∴ 2x³ – x² – 13x – 6 = (2x + 1)(x² – x – 6)

So, one factor of the given expression is (2x + 1). Now, we need to factorize (x² – x – 6).

For factorizing, this expression, split the middle term in such a way that the product of the coefficients of the new terms is equal to the product of the coefficients of the first and last terms in the expression.

Here, product of co-effs of first and last terms = 1 × (–6) = –6

So, if the middle term –x is split into two terms say ax, bx, then a + b = –1 and ab = –6.

Observe that values –3 and 2 satisfy these equations.

⇒ x² – x – 6 = x² – 3x + 2x – 6

Observe that x is common for the first two terms and 2 is common for the next two terms.

⇒ x² – 3x + 2x – 6 = x(x – 3) + 2(x – 3)

Now, (x – 3) is the common term.

⇒ x² – x – 6 = (x – 3)(x + 2)

Thus, the factors of x² – x – 6 are (x – 3) and (x + 2).

∴ 2x³ – x² – 13x – 6 = (2x + 1)(x – 3)(x + 2)

Hence, the factors of 2x³ – x² – 13x – 6 are (2x + 1), (x – 3) and (x + 2).