Answer :

6x^{2} + 7x – 3

The product of 1^{st} and the 3^{rd} term i.e. 18 can be factorized such that the sum or difference should be equal to the middle term.

Thus, the factors of 18 are 2 and 7

⇒ 6x^{2} + 7x – 3

⇒ 6x^{2} + 9x – 2x – 3

⇒ 3x(2+3) – 1(2x+3)

⇒ (3x – 1) (2x+3) Ans.

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Factorize:

x^{3} – 2x^{2}y + 3xy^{2} – 6y^{3}

Find the value of a for which the polynomial is divisible by (x+3).

RS Aggarwal & V Aggarwal - MathematicsFactorize:

a^{2} + a – 3a^{2} - 3

Factorize:

2a(x + y) -3b(x + y)

RS Aggarwal & V Aggarwal - MathematicsFactorize:

8 – 4a – 2a^{3} + a^{4}

If 𝑎 + 𝑏 + 𝑐 = 9 and 𝑎𝑏 + 𝑏𝑐 + 𝑐𝑎 = 26, find 𝑎^{2} + 𝑏^{2} + 𝑐^{2}.

Show that p – 1 is a factor of p^{10} – 1 and also of p^{11} – 1.

Factorize:

a^{2} + ab(b + 1) + b^{3}

Find the value of m so that 2x – 1 be a factor of 8𝑥^{4} + 4𝑥^{3}−16𝑥^{2} + 10𝑥 + 𝑚.