A number puzzle: the polynomial p(x) = x^3 + 13x^2 + 32x + 20 has all three roots negative integers. Given (x + 1) is one factor, and the quotient is x^2 + 12x + 20, factorise p(x) completely.

Question

Accepted Answer

(x + 1)(x + 2)(x + 10). x^2 + 12x + 20 = (x + 2)(x + 10) since 2 + 10 = 12 and 2 x 10 = 20. So p(x) = (x + 1)(x + 2)(x + 10).

A number puzzle: the polynomial $p (x) = x^{3} + 13 x^{2} + 32 x + 20$ has all three roots negative integers.
Given $(x + 1)$ is one factor, and the quotient is $x^{2} + 12 x + 20$ , factorise $p (x)$ completely.

Step-by-Step Solution

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More Class 9 Factor Theorem and Factorisation practice

A number puzzle: the polynomial p(x)=x3+13x2+32x+20 has all three roots negative integers.Given (x+1) is one factor, and the quotient is x2+12x+20, 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 number puzzle: the polynomial $p (x) = x^{3} + 13 x^{2} + 32 x + 20$ has all three roots negative integers.
Given $(x + 1)$ is one factor, and the quotient is $x^{2} + 12 x + 20$ , factorise $p (x)$ completely.