Question

If \[p\left( x \right)={{x}^{3}}+2{{x}^{2}}-5x-6.\] Find \[p\left( 2 \right),\text{ }p\left( -1 \right),\text{ }p\left( -3 \right)\] and \[p\left( 0 \right).\] What can you say about zero of \[p\left( x \right).\]

Answer 1

Hint: We know that the number of zeros solutions of an expression or an equation depends upon the highest degree of its variable. If the highest degree is two, then that expression or equation will have exactly two solutions; if the highest degree is four, then it will have exactly four solutions.

Complete step by step solution:
As we know the Zeros or roots of a polynomial: For a polynomial, there could be some values of the variable for which the polynomial will be zero. These values are called zeros of a polynomial. We can also call these as roots of the polynomials. To find the zeros of a polynomial, we equate it to zero and find the value of the variable. Thus, the highest degree of x is \[3\] so the above expression had exactly $3$ solutions. Substituting the value of x as zero will not give the expression value as zero always
We are given to find the values \[p\left( 2 \right),\text{ }p\left( -1 \right),\text{ }p\left( -3 \right)~~\And ~~p\left( 0 \right)\] and the zeroes of the expression \[p\left( x \right)={{x}^{3}}+2{{x}^{2}}-5x-6.\]

1. For finding the value of \[p\left( 2 \right),\] we are substituting $2$ in the place of x in $p\left( x \right)$
$p\left( 2 \right)={{\left( 2 \right)}^{3}}+2{{\left( 2 \right)}^{2}}-5\left( 2 \right)-6$
On further solving we get; $ \left[ 8+8-10-6 \right]=16-16=0$

2. For finding the value of \[\text{ }p\left( -1 \right),\] we are substituting $-1$ in the place of x in $p\left( x \right)$
$p\left( -1 \right)={{\left( -1 \right)}^{3}}+2{{\left( -1 \right)}^{2}}-5\left( -1 \right)-6$
On further solving we get; $ \left[ -1+2+5-6 \right]=7-7=0$

3. For finding the value of \[p\left( -3 \right)~\] we are substituting $-3$ in the place of x in $p\left( x \right)$
$p\left( 3 \right)={{\left( -3 \right)}^{3}}+2{{\left( -3 \right)}^{2}}-5\left( -3 \right)-6$
On further solving we get; $ \left[ -27+18+15-6 \right]=33-33=0$

4. For finding the value of \[p\left( 0 \right)\] we are substituting $0$ in the place of x in $p\left( x \right)$
\[p\left( 0 \right)={{\left( 0 \right)}^{3}}+2{{\left( 0 \right)}^{2}}-5\left( 0 \right)-6\]
On further solving we get; \[\left[ 0+0-0-6 \right]=-6\]
So do not be confused that the number \[0\] is not a zero solution of an equation always. The zero of polynomials can be defined as those values of x when substituted in the polynomial, making it equal to zero. In other words, we can say that the zeroes are the roots of the polynomial. We can obtain the zeroes of the polynomial P(x) by equating it to zero.
Hence $p\left( 2 \right)=0,p\left( -1 \right)=0~~\And ~~p\left( -3 \right)=0$ so that $2,-1~~\And ~~3$ are zeros of \[p\left( x \right),\] where $p\left( 0 \right)$ is not zero of \[p\left( x \right)\] since $p\left( 0 \right)=-6\ne 0.$

Note:
Remember that while solving the equations and take care of the sign while taking the terms to the right-hand side. Also remember that the number of zeroes of the polynomial is equal to the maximum exponent of the variable in the polynomial.