Question

(a). Check whether $\left( {x - 1} \right)$ is a factor of the polynomial $p\left( x \right) = 6{x^3} + 3{x^2}$
(b). What first degree polynomial added to $p\left( x \right)$ gives a polynomial for which $\left( {{x^2} - 1} \right)$ is a factor?

Answer 1

Hint:
(a). To check whether \[\left( {x - 1} \right)\] is a factor or not we have to put the value of x as 1 and if \[x = 1\] satisfies the given equation then \[\left( {x - 1} \right)\] is a factor of given equation.
(b). Let \[f\left( x \right)\] be the required polynomial. Then, we will assume the first-degree polynomial as $ax + b$
After that, If ${x^2} - 1$ is a factor of the $f\left( x \right)$ , then $f\left( { - 1} \right) = 0$ and $f\left( 1 \right) = 0$
Then, we will put the value of $f\left( { - 1} \right)$ and $f\left( 1 \right)$ in $f\left( x \right) = p\left( x \right) + ax + b$ and then after we will get one equation from $f\left( { - 1} \right)$ and another equation from $f\left( 1 \right)$ .
Finally, after comparing the equation of $f\left( { - 1} \right)$ and $f\left( 1 \right)$ we will get the answer.

Complete step by step solution:
(a). Here, we have to find if $\left( {x - 1} \right)$ is a factor of $p\left( x \right) = 6{x^3} + 3{x^2}$
If, $\left( {x - 1} \right)$ is a factor of $p\left( x \right) = 6{x^3} + 3{x^2}$ then when we put $x - 1 = 0 \Rightarrow x = 1$ in the given polynomial then it must give the value as 0.
Thus, $p\left( 1 \right) = 0$ if $\left( {x - 1} \right)$ is a factor of $p\left( x \right) = 6{x^3} + 3{x^2}$ .
Now, put the value of \[x = 1\] in equation
 $
\Rightarrow p\left( 1 \right) = 6{\left( 1 \right)^3} + 3{\left( 1 \right)^2} \\
 \Rightarrow p\left( 1 \right) = 6 + 3 \\
 \Rightarrow p\left( 1 \right) = 9 \ne 0 \\
 $
Hence, $\left( {x - 1} \right)$ is not a factor of $p\left( x \right) = 6{x^3} + 3{x^2}$ .
(b). Let f(x) be the required polynomial,
 $p\left( x \right) = 6{x^3} + 3{x^2}$
Let as assume the first-degree polynomial as $ax + b$
 $f\left( x \right) = p\left( x \right) + ax + b$
If ${x^2} - 1$ is a factor of the $f\left( x \right)$ , then $f\left( { - 1} \right) = 0$ and $f\left( 1 \right) = 0$
 $f\left( { - 1} \right) = p\left( { - 1} \right) + a\left( { - 1} \right) + b$
 \[ \Rightarrow 0 = 6{\left( { - 1} \right)^3} + 3{\left( { - 1} \right)^2} - a + b\]
 $ \Rightarrow 0 = - 6 + 3 - a + b$
 $ \Rightarrow a - b = - 3$ (I)
Also, $f\left( 1 \right) = p\left( 1 \right) + a\left( 1 \right) + b$
 $ \Rightarrow 0 = - 6 + 3 - a + b$
 $ \Rightarrow a - b = - 3$
Solving (I) and (II), we get
 $a = - 6, b = - 3$

\[ \Rightarrow \] The polynomial is $\left( { - 6x, - 3} \right)$

Note:
Polynomial: Polynomials are the algebraic expressions which consist of variables and coefficient. Variables are also sometimes called indeterminants. We can perform the arithmetic operations such as addition, subtraction, multiplication, and also positive integer exponents for polynomial expressions but not division by variable.
The word polynomial is derived from the Greek word ‘poly’ means “many” and ‘nominal’ means “terms”, so altogether is said “many terms”.