Question 2 and 3, Exercise 5.1

Solutions of Question 2 and 3 of Exercise 5.1 of Unit 05: Polynomials. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Show that $x-3$ is a factor of $x^{3}-2 x^{2}-5 x+6$.

Solution.

Let $p(x)=x^{3}-2 x^{2}-5 x+6$ and $x-c=x-3$ $\implies c=3$.

By factor theorem $x-3$ is factor of $p(x)$ iff $p(3)=0$.

Now $$\begin{align*} p(3)&=3^3-2(3)^2-5(3)+6 \\ & = 27-18-15+6 \\ & = 0. \end{align*}$$ Hence, $x-3$ is factor of $p(x)$.

Decide whether $x-3$ is a factor of $x^{3}-2 x^{2}-5 x+1$ or not.

Solution.

Let $p(x)=x^{3}-2 x^{2}-5 x+1$ and $x-c=x-3$ $\implies c=3$.

By factor theorem $x-3$ is factor of $p(x)$ iff $p(3)=0$.

Now $$\begin{align*} p(3)&=3^3-2(3)^2-5(3)+1 \\ & = 27-18-15+1 \\ & = -5 \neq 0 \end{align*}$$ Hence, $x-3$ is not factor of $p(x)$.

< Question 1 Question 4 & 5 >