A cubic p(x) = 2x^3 + 3x^2 - 11x - 6 has (x - 2) as a factor (since p(2) = 0). Dividing by (x - 2) gives the quotient 2x^2 + 7x + 3. Factorise p(x) completely.

Question

Accepted Answer

(x - 2)(2x + 1)(x + 3). 2x^2 + 7x + 3 = (2x + 1)(x + 3). So p(x) = (x - 2)(2x + 1)(x + 3). Check p(2)=16+12-22-6=0.

A cubic $p (x) = 2 x^{3} + 3 x^{2} - 11 x - 6$ has $(x - 2)$ as a factor (since $p (2) = 0$ ).
Dividing by $(x - 2)$ gives the quotient $2 x^{2} + 7 x + 3$ .
Factorise $p (x)$ completely.

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More Class 9 Factor Theorem and Factorisation practice

A cubic p(x)=2x3+3x2−11x−6 has (x−2) as a factor (since p(2)=0).Dividing by (x−2) gives the quotient 2x2+7x+3.Factorise p(x) completely.

Step-by-Step Solution

Master this chapter in 30 minutes a day.

More Class 9 Factor Theorem and Factorisation practice

A cubic $p (x) = 2 x^{3} + 3 x^{2} - 11 x - 6$ has $(x - 2)$ as a factor (since $p (2) = 0$ ).
Dividing by $(x - 2)$ gives the quotient $2 x^{2} + 7 x + 3$ .
Factorise $p (x)$ completely.