Question

How do you find all the zeros of \[P\left( x \right)={{x}^{4}}+{{x}^{3}}-2{{x}^{2}}+4x-24\]?

Answer 1

Hint: This question is from the topic of algebra. In this question, we will first understand the rational root theorem. After that, we will find the first zero of the equation \[P\left( x \right)={{x}^{4}}+{{x}^{3}}-2{{x}^{2}}+4x-24\] using the rational root theorem. After that, we will find the other roots of the equation by factoring the equation. After solving, we will get all the zeros of the equation.

Complete step by step solution:
Let us solve this question.
In this question, it is asking to find out the zeroes of the equation \[P\left( x \right)={{x}^{4}}+{{x}^{3}}-2{{x}^{2}}+4x-24\]
For finding out the zeroes of the equation, we will use rational root theorem.
The rational root theorem says that the zeros or the roots of the any given function or equation \[y={{a}_{0}}{{x}^{n}}+{{a}_{1}}{{x}^{n-1}}+{{a}_{2}}{{x}^{n-2}}.............+{{a}_{n-1}}x+{{a}_{n}}\] is expressed in the form of \[\dfrac{p}{q}\], where the value of p are the factors of the term \[{{a}_{n}}\] and the values of q are the factors of the term \[{{a}_{0}}\].
So, the zeros of the equation \[P\left( x \right)={{x}^{4}}+{{x}^{3}}-2{{x}^{2}}+4x-24\] will be in the form of \[\dfrac{p}{q}\], where value of p is a factor of -24 and q is a factor of 1.
Hence, the possible zeros of the equation \[P\left( x \right)={{x}^{4}}+{{x}^{3}}-2{{x}^{2}}+4x-24\] will be
\[\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 8,\pm 12,\pm 24\]
Now, we will check the value of equation \[P\left( x \right)={{x}^{4}}+{{x}^{3}}-2{{x}^{2}}+4x-24\] at the above values by putting those values in place of x.
If the value of \[P\left( x \right)\] is zero at some value of x, then that value of x will be zero or root of the equation.
Let us check the value of equation \[P\left( x \right)={{x}^{4}}+{{x}^{3}}-2{{x}^{2}}+4x-24\] at x=1
\[P\left( 1 \right)={{1}^{4}}+{{1}^{3}}-2\times {{1}^{2}}+4\times 1-24=1+1-2+4-24=-20\]
Hence, x=1 is not a zero of the equation.
Let us check at x=-1.
\[P\left( -1 \right)={{\left( -1 \right)}^{4}}+{{\left( -1 \right)}^{3}}-2\times {{\left( -1 \right)}^{2}}+4\times \left( -1 \right)-24=1-1-2-4-24=-30\]
Hence, x=-1 is not a zero of the equation.
Let us check at x=2.
\[P\left( 2 \right)={{\left( 2 \right)}^{4}}+{{\left( 2 \right)}^{3}}-2\times {{\left( 2 \right)}^{2}}+4\times \left( 2 \right)-24=16+8-8+8-24=0\]
Hence, we get that x=2 is a zero of the equation. So, (x-2) will be a factor of the factor of the equation \[P\left( x \right)={{x}^{4}}+{{x}^{3}}-2{{x}^{2}}+4x-24\].
So, we can write
\[P\left( x \right)=\left( x-2 \right)\left( {{x}^{3}}+3{{x}^{2}}+4x+12 \right)\]
Now, we have to find out the factor of the equation \[\left( {{x}^{3}}+3{{x}^{2}}+4x+12 \right)\]
Here, in the equation \[\left( {{x}^{3}}+3{{x}^{2}}+4x+12 \right)\], we can see that in the first two terms after taking common as \[{{x}^{2}}\], then we will get \[{{x}^{2}}\] multiplied by (x+3) and similarly if we take 4 common in the last two terms then we will get 4 multiplied by (x+3).
Hence, we can write the equation \[\left( {{x}^{3}}+3{{x}^{2}}+4x+12 \right)\] as \[{{x}^{2}}\left( x+3 \right)+4\left( x+3 \right)\] which can also be written as \[\left( x+3 \right)\left( {{x}^{2}}+4 \right)\].
Therefore, we can write the equation \[{{x}^{4}}+{{x}^{3}}-2{{x}^{2}}+4x-24\] as \[\left( x-2 \right)\left( x+3 \right)\left( {{x}^{2}}+4 \right)\].
Hence, the zeros will be
\[x=2\], \[x=-3\], and \[{{x}^{2}}=-4\] that is \[x=\sqrt{-4}=2i\]
So, we have found all the zeros of the equation \[P\left( x \right)={{x}^{4}}+{{x}^{3}}-2{{x}^{2}}+4x-24\] and that are 2, -3 and \[2i\].

Note:
As we can see that this question belongs to the topic of algebra, so we should have a better knowledge in that topic. We should know that a square of negative of 1 is always iota. The symbol of iota is \[i\]. We should know about rational root theorems. If a equation is given as \[y={{a}_{0}}{{x}^{n}}+{{a}_{1}}{{x}^{n-1}}+{{a}_{2}}{{x}^{n-2}}.............+{{a}_{n-1}}x+{{a}_{n}}\], then the roots of the equation is always in the form of \[\dfrac{p}{q}\], where p are the factors of the term \[{{a}_{n}}\] and the values of q are the factors of the term \[{{a}_{0}}\].