
How do you factor the polynomial function $y = 2{x^3} - 5{x^2} + 4$?
Answer
531k+ views
Hint: In this question, we need to find the factors of the given polynomial. Firstly, we put different values of x in the given equation until the value of the equation becomes zero. The value of x which makes the equation zero is one of the factors. Then we write the given polynomial in such a way that the obtained factor before turns out as a common factor. Then we simplify it further to obtain the remaining factors.
Complete step-by-step solution:
Given the polynomial of the form $y = 2{x^3} - 5{x^2} + 4$
We are asked to find the factors of the above polynomial.
Note that the degree of the polynomial is 3.
Let us take the polynomial as some $h(x)$.
i.e. $h(x) = y = 2{x^3} - 5{x^2} + 4$ …… (1)
We have to find a value of the variable x which will make the whole expression equal to zero.
The value of x which makes the equation zero will be one of the factors.
We put different values of x in the equation (1).
Let $x = 1$, we have,
$ \Rightarrow h(1) = 2{(1)^3} - 5{(1)^2} + 4$
$ \Rightarrow h(1) = 2 - 5 + 4$
$ \Rightarrow h(1) = 1 \ne 0$
So, $(x - 1)$ is not a factor.
Let $x = - 1$, we have,
$ \Rightarrow h( - 1) = 2{( - 1)^3} - 5{( - 1)^2} + 4$
$ \Rightarrow h( - 1) = - 2 - 5 + 4$
$ \Rightarrow h( - 1) = - 3 \ne 0$
So, $(x + 1)$ is not a factor.
Let $x = 2$, we have,
$ \Rightarrow h(2) = 2{(2)^3} - 5{(2)^2} + 4$
$ \Rightarrow h(2) = 16 - 20 + 4$
$ \Rightarrow h(2) = 0$
So, $(x - 2)$ is a factor.
Now we arrange the terms of $h(x)$ in such a way that $(x - 2)$ turns out as a common factor.
Now we write $h(x)$ as,
$h(x) = 2{x^3} - 5{x^2} + 4$
This can also be written as,
$ \Rightarrow 2{x^3} - 4{x^2} - {x^2} + 4$
Add and subtract $2x$, we get,
$ \Rightarrow 2{x^3} - 4{x^2} - {x^2} + 2x - 2x + 4$
Factor out the common terms, we get,
$ \Rightarrow 2{x^2}(x - 2) - x(x - 2) - 2(x - 2)$
Now factoring out the common factor, we get,
$ \Rightarrow (x - 2)(2{x^2} - x - 2)$
Hence the factors of the polynomial $y = 2{x^3} - 5{x^2} + 4$ is given by $(x - 2)(2{x^2} - x - 2)$.
Note: Students need to be careful in solving such problems. They need to choose the first factor very carefully as the other factors are determined on its basis only. So substitute different values for the unknown variable and simplify so that the equation becomes zero. The value of the variable for which the equation becomes zero is one of the required factors. Later on we need to simplify the polynomial in such a way that the obtained factor must turn out as a common factor. This is one of the important steps to be remembered.
Complete step-by-step solution:
Given the polynomial of the form $y = 2{x^3} - 5{x^2} + 4$
We are asked to find the factors of the above polynomial.
Note that the degree of the polynomial is 3.
Let us take the polynomial as some $h(x)$.
i.e. $h(x) = y = 2{x^3} - 5{x^2} + 4$ …… (1)
We have to find a value of the variable x which will make the whole expression equal to zero.
The value of x which makes the equation zero will be one of the factors.
We put different values of x in the equation (1).
Let $x = 1$, we have,
$ \Rightarrow h(1) = 2{(1)^3} - 5{(1)^2} + 4$
$ \Rightarrow h(1) = 2 - 5 + 4$
$ \Rightarrow h(1) = 1 \ne 0$
So, $(x - 1)$ is not a factor.
Let $x = - 1$, we have,
$ \Rightarrow h( - 1) = 2{( - 1)^3} - 5{( - 1)^2} + 4$
$ \Rightarrow h( - 1) = - 2 - 5 + 4$
$ \Rightarrow h( - 1) = - 3 \ne 0$
So, $(x + 1)$ is not a factor.
Let $x = 2$, we have,
$ \Rightarrow h(2) = 2{(2)^3} - 5{(2)^2} + 4$
$ \Rightarrow h(2) = 16 - 20 + 4$
$ \Rightarrow h(2) = 0$
So, $(x - 2)$ is a factor.
Now we arrange the terms of $h(x)$ in such a way that $(x - 2)$ turns out as a common factor.
Now we write $h(x)$ as,
$h(x) = 2{x^3} - 5{x^2} + 4$
This can also be written as,
$ \Rightarrow 2{x^3} - 4{x^2} - {x^2} + 4$
Add and subtract $2x$, we get,
$ \Rightarrow 2{x^3} - 4{x^2} - {x^2} + 2x - 2x + 4$
Factor out the common terms, we get,
$ \Rightarrow 2{x^2}(x - 2) - x(x - 2) - 2(x - 2)$
Now factoring out the common factor, we get,
$ \Rightarrow (x - 2)(2{x^2} - x - 2)$
Hence the factors of the polynomial $y = 2{x^3} - 5{x^2} + 4$ is given by $(x - 2)(2{x^2} - x - 2)$.
Note: Students need to be careful in solving such problems. They need to choose the first factor very carefully as the other factors are determined on its basis only. So substitute different values for the unknown variable and simplify so that the equation becomes zero. The value of the variable for which the equation becomes zero is one of the required factors. Later on we need to simplify the polynomial in such a way that the obtained factor must turn out as a common factor. This is one of the important steps to be remembered.
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