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How do you factor \[y={{x}^{3}}+4{{x}^{2}}-x-4\]?

seo-qna
Last updated date: 21st Jul 2024
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Views today: 10.83k
Answer
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Hint: This question is from the topic of algebra. We will factor the equation \[y={{x}^{3}}+4{{x}^{2}}-x-4\] by using rational root theorem. According to rational root theorem, first we will check the possibilities of roots of the equation. After getting the exact zeros, we will find the factors of the equation \[y={{x}^{3}}+4{{x}^{2}}-x-4\].

Complete step by step answer:
Let us solve this question.
The question is asking us to factor the given equation. The given equation is \[y={{x}^{3}}+4{{x}^{2}}-x-4\].
For finding out the factor of the equation, we will use rational root theorem.
The rational root theorem says that the zeros or the roots of any given function or equation is expressed in the form of \[\dfrac{p}{q}\], where p and q are integers. If we consider an example of an equation like:
\[y={{a}_{0}}{{x}^{n}}+{{a}_{1}}{{x}^{n-1}}+{{a}_{2}}{{x}^{n-2}}.............+{{a}_{n-1}}x+{{a}_{n}}\],
Then we can say that the values of p are factors of \[{{a}_{n}}\] and the values of q are the factors of \[{{a}_{0}}\] or we can say that p should be a divisor of constant term and q should be a divisor of coefficient of leading term.
So, the zeros of the equation \[y={{x}^{3}}+4{{x}^{2}}-x-4\] will be in the form of \[\dfrac{p}{q}\], where value of p is a factor of -4 and value of q is a factor of 1.
Hence, possibilities of zeros of the equation \[y={{x}^{3}}+4{{x}^{2}}-x-4\] will be
\[\pm 1\], \[\pm 2\], and \[\pm 4\].
Now, we will check the value of equation \[y={{x}^{3}}+4{{x}^{2}}-x-4\] by putting the above values in place of x.
If the value of y is zero at any value, then it will be the root of the equation. And, if the value of y is not zero, then the term which we have put in place of x will not be the root of the equation.
Let us check the value of equation \[y={{x}^{3}}+4{{x}^{2}}-x-4\] at x=-1.
\[y={{\left( -1 \right)}^{3}}+4{{\left( -1 \right)}^{2}}-\left( -1 \right)-4=-1+4+1-4=0\]
As the value of y is zero at x=-1. Then, -1 is a root or zero of the equation.
Now, let us check at x=1, we get
\[y={{\left( 1 \right)}^{3}}+4{{\left( 1 \right)}^{2}}-\left( 1 \right)-4=1+4-1-4=0\]
Hence, we get that x=+1 is a root of the equation.
Now, let us check at x=-2, we get
\[y={{\left( -2 \right)}^{3}}+4{{\left( -2 \right)}^{2}}-\left( -2 \right)-4=-8+16+2-4=6\]
Hence, x=-2 is not the root of the equation.
Let us check at x=+2, we get
\[y={{\left( +2 \right)}^{3}}+4{{\left( +2 \right)}^{2}}-\left( +2 \right)-4=8+16-2-4=18\]
Hence, x=+2 is not a root of the equation.
Now, let us check at x=-4, we get
\[y={{\left( -4 \right)}^{3}}+4{{\left( -4 \right)}^{2}}-\left( -4 \right)-4=-64+64+4-4=0\]
Hence, we get that x=-4 is the root of the equation.
Now, let us check at x=+4, we get
\[y={{\left( +4 \right)}^{3}}+4{{\left( +4 \right)}^{2}}-\left( +4 \right)-4=64+64-4-4=120\]
Hence, x=+4 is not the root of the equation.
Now, we have seen that -1, +1, and -4 is a root of the equation.
As x=-1 is root of the equation \[y={{x}^{3}}+4{{x}^{2}}-x-4\], then we can say that (x+1) is a factor of the equation \[y={{x}^{3}}+4{{x}^{2}}-x-4\].
 Similarly, we can say that if x=+1 and x=-4 are roots, then (x-1) and (x+4) are factors of the equation.
So, we get that (x+1), (x-1), and (x+4) are the factors of the equation \[y={{x}^{3}}+4{{x}^{2}}-x-4\].

Therefore, we can write the factors of equation as
\[y={{x}^{3}}+4{{x}^{2}}-x-4=\left( x+1 \right)\left( x-1 \right)\left( x+4 \right)\].


Note: We should have a better knowledge in algebra and polynomials to solve this type of question easily. We should know about the rational root theorem. As we have used rational root theorem in this question. This theorem can be used in any type of polynomial rather it satisfies the conditions. And, also remember that if any equation \[y={{a}_{0}}{{x}^{n}}+{{a}_{1}}{{x}^{n-1}}+{{a}_{2}}{{x}^{n-2}}.............+{{a}_{n-1}}x+{{a}_{n}}\] (lets, say) has a root that is \[{{x}_{1}}\], then \[\left( x-{{x}_{1}} \right)\] will be a factor of the equation \[y={{a}_{0}}{{x}^{n}}+{{a}_{1}}{{x}^{n-1}}+{{a}_{2}}{{x}^{n-2}}.............+{{a}_{n-1}}x+{{a}_{n}}\]