Question

How do you find all the zeros of \[f\left( x \right) = {x^4} + 6{x^3} + 14{x^2} + 54x + 45\]?

Answer 1

Hint: Here in this question, we have to find the zeros of the given polynomial \[f\left( x \right)\]. To solve this, we have to find the value \[x\], when substitute the \[x\] value in polynomial \[f\left( x \right)\] after simplification the result should be zero. When the whole polynomial is zero, that \[x\] value is the zeros of the given polynomial.

Complete step by step answer:
we say that a zero of a polynomial \[p\left( x \right)\] is a number \[c\] such that \[p\left( c \right) = 0\]. You must have observed that the zero of the polynomial \[x - 1\] is obtained by equating it to \[0\], i.e., \[x - 1 = 0\], which gives \[x = 1\]. We say \[p\left( x \right) = 0\] is a polynomial equation and 1 is the root of the polynomial equation \[p\left( x \right) = 0\].
Some observations of zeros of polynomials is:
A zero of a polynomial need not be 0.
 0 may be a zero of a polynomial.
Every linear polynomial has one and only one zero.
 A polynomial can have more than one zero.
Consider the given polynomial
\[f\left( x \right) = {x^4} + 6{x^3} + 14{x^2} + 54x + 45\]--------(1)
By guess working when \[x = - 1\], then
\[ \Rightarrow \,\,f\left( { - 1} \right) = {\left( { - 1} \right)^4} + 6{\left( { - 1} \right)^3} + 14{\left( { - 1} \right)^2} + 54\left( { - 1} \right) + 45\]
\[ \Rightarrow \,\,f\left( { - 1} \right) = 1 - 6 + 14 - 54 + 45\]
\[ \Rightarrow \,\,f\left( { - 1} \right) = 60 - 60\]
\[ \Rightarrow \,\,f\left( { - 1} \right) = 0\]
So \[x = - 1\]is a zero and \[\left( {x + 1} \right)\] a factor:
Then equation (1) can be written as
\[f\left( x \right) = {x^4} + {x^3} + 5{x^3} + 5{x^2} + 9{x^2} + 9x + 45x + 45\]
\[f\left( x \right) = {x^4} + {x^3} + 5{x^3} + 5{x^2} + 9{x^2} + 9x + 45x + 45\]
Take out greatest common divisor
\[f\left( x \right) = {x^3}\left( {x + 1} \right) + 5{x^2}\left( {x + 1} \right) + 9x\left( {x + 1} \right) + 45\left( {x + 1} \right)\]
Take \[\left( {x + 1} \right)\] common
\[f\left( x \right) = \left( {x + 1} \right)\left( {{x^3} + 5{x^2} + 9x + 45} \right)\]
Looking at the remaining cubic, note that the ratio between the first and second terms is the same as that between the third and fourth terms. So this cubic will factor by grouping.
Now consider the cubic equation
\[{x^3} + 5{x^2} + 9x + 45\]
Now, grouping the factors
\[ \Rightarrow \,\,\left( {{x^3} + 5{x^2}} \right) + \left( {9x + 45} \right)\]
\[ \Rightarrow \,\,{x^2}\left( {x + 5} \right) + 9\left( {x + 5} \right)\]
Take \[\left( {x + 5} \right)\] common, then
\[ \Rightarrow \,\,\left( {x + 5} \right)\left( {{x^2} + 9} \right)\]
Now, equate the each factors to zero, then
\[ \Rightarrow \,\,x + 5 = 0\] or \[{x^2} + 9 = 0\]
\[ \Rightarrow \,\,x = - 5\] or \[x = \pm \sqrt { - 9} = \pm 3i\]

Hence, the zeros of the given polynomial \[f\left( x \right) = {x^4} + 6{x^3} + 14{x^2} + 54x + 45\] are \[-1,\, - 5,\,\]and \[ \pm 3\,i\]

Note: 1.The zeros of any polynomial equation will be the value of the x term. we have to find the value of x and if we substitute the value of x in the given equation and hence we obtain the solution is zero. The number of roots will depend on the degree of the polynomial equation.