Priya tests several three-digit candidate roots and finds that x = 3 makes p(x) = x^3 - 4x^2 + x + 6 equal to zero. This means (x - 3) is a factor. After dividing, the quotient is x^2 - x - 2. Factorise p(x) fully.

Question

Accepted Answer

(x - 3)(x - 2)(x + 1). x^2 - x - 2 = (x - 2)(x + 1). So p(x) = (x - 3)(x - 2)(x + 1). Check p(2)=8-16+2+6=0, p(-1)=-1-4-1+6=0.

Priya tests several three-digit candidate roots and finds that $x = 3$ makes $p (x) = x^{3} - 4 x^{2} + x + 6$ equal to zero.
This means $(x - 3)$ is a factor.
After dividing, the quotient is $x^{2} - x - 2$ . Factorise $p (x)$ fully.

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More Class 9 Factor Theorem and Factorisation practice

Priya tests several three-digit candidate roots and finds that x=3 makes p(x)=x3−4x2+x+6 equal to zero.This means (x−3) is a factor.After dividing, the quotient is x2−x−2. Factorise p(x) fully.

Step-by-Step Solution

Master this chapter in 30 minutes a day.

More Class 9 Factor Theorem and Factorisation practice

Priya tests several three-digit candidate roots and finds that $x = 3$ makes $p (x) = x^{3} - 4 x^{2} + x + 6$ equal to zero.
This means $(x - 3)$ is a factor.
After dividing, the quotient is $x^{2} - x - 2$ . Factorise $p (x)$ fully.