
How do you factor $3{y^4} - 2{y^2} - 5$ completely?
Answer
545.1k+ views
Hint: To solve this question, we will assume$x = {y^2}$. Now, this equation is the quadratic equation. The general form of the quadratic equation is$a{x^2} + bx + c = 0$. Where ‘a’ is the coefficient of${x^2}$, ‘b’ is the coefficient of x and ‘c’ is the constant term. To solve this equation, we will apply the sum-product pattern. During the simplification, we will take out common factors from the two pairs. Then we will rewrite it in factored form.
Therefore, we should follow the below steps:
> Apply sum-product pattern.
> Make two pairs.
> Common factor from two pairs.
> Rewrite in factored form.
Complete step-by-step answer:
In this question, we want to factor in the given expression.
The given expression is,
$ \Rightarrow 3{y^4} - 2{y^2} - 5$
In the above expression, we can write the term ${y^4}$as ${\left( {{y^2}} \right)^2}$ .
Let us substitute ${y^4}$term in the above expression.
$ \Rightarrow 3{\left( {{y^2}} \right)^2} - 2{y^2} - 5$
Now, let us assume$x = {y^2}$.
Substitute x in the above expression instead of ${y^2}$.
$ \Rightarrow 3{x^2} - 2x - 5$
This equation is the quadratic equation.
Let us apply the sum-product pattern in the above equation.
Since the coefficient of ${x^2}$is 3 and the constant term is -5. Let us multiply 3 and -5. The answer will be -15. We have to find the factors of 15 which sum to -2. Here, the factors are -5 and 3.
Therefore,
$ \Rightarrow 3{x^2} - 5x + 3x - 5$
Now, make two pairs in the above equation.
$ \Rightarrow \left( {3{x^2} - 5x} \right) + \left( {3x - 5} \right)$
Let us take out the common factor.
$ \Rightarrow x\left( {3x - 5} \right) + 1\left( {3x - 5} \right)$
Now, rewrite the above equation in factored form.
$ \Rightarrow \left( {3x - 5} \right)\left( {x + 1} \right)$
Now, substitute ${y^2}$ in the above expression instead of x.
That is equal to,
$ \Rightarrow \left( {3{y^2} - 5} \right)\left( {{y^2} + 1} \right)$
Hence, the factors are $\left( {3{y^2} - 5} \right)$and $\left( {{y^2} + 1} \right)$.
Note:
One important thing is, we can always check our work by multiplying out factors back together, and check that we have got back the original answer.
To check our factorization, multiplication goes like this:
$ \Rightarrow \left( {3x - 5} \right)\left( {x + 1} \right)$
Let us apply multiplication to remove brackets.
$ \Rightarrow 3{x^2} - 5x + 3x - 5$
Let us simplify it. We will get,
$ \Rightarrow 3{x^2} - 2x - 5$
Hence, we get our quadratic equation back by applying multiplication.
Therefore, we should follow the below steps:
> Apply sum-product pattern.
> Make two pairs.
> Common factor from two pairs.
> Rewrite in factored form.
Complete step-by-step answer:
In this question, we want to factor in the given expression.
The given expression is,
$ \Rightarrow 3{y^4} - 2{y^2} - 5$
In the above expression, we can write the term ${y^4}$as ${\left( {{y^2}} \right)^2}$ .
Let us substitute ${y^4}$term in the above expression.
$ \Rightarrow 3{\left( {{y^2}} \right)^2} - 2{y^2} - 5$
Now, let us assume$x = {y^2}$.
Substitute x in the above expression instead of ${y^2}$.
$ \Rightarrow 3{x^2} - 2x - 5$
This equation is the quadratic equation.
Let us apply the sum-product pattern in the above equation.
Since the coefficient of ${x^2}$is 3 and the constant term is -5. Let us multiply 3 and -5. The answer will be -15. We have to find the factors of 15 which sum to -2. Here, the factors are -5 and 3.
Therefore,
$ \Rightarrow 3{x^2} - 5x + 3x - 5$
Now, make two pairs in the above equation.
$ \Rightarrow \left( {3{x^2} - 5x} \right) + \left( {3x - 5} \right)$
Let us take out the common factor.
$ \Rightarrow x\left( {3x - 5} \right) + 1\left( {3x - 5} \right)$
Now, rewrite the above equation in factored form.
$ \Rightarrow \left( {3x - 5} \right)\left( {x + 1} \right)$
Now, substitute ${y^2}$ in the above expression instead of x.
That is equal to,
$ \Rightarrow \left( {3{y^2} - 5} \right)\left( {{y^2} + 1} \right)$
Hence, the factors are $\left( {3{y^2} - 5} \right)$and $\left( {{y^2} + 1} \right)$.
Note:
One important thing is, we can always check our work by multiplying out factors back together, and check that we have got back the original answer.
To check our factorization, multiplication goes like this:
$ \Rightarrow \left( {3x - 5} \right)\left( {x + 1} \right)$
Let us apply multiplication to remove brackets.
$ \Rightarrow 3{x^2} - 5x + 3x - 5$
Let us simplify it. We will get,
$ \Rightarrow 3{x^2} - 2x - 5$
Hence, we get our quadratic equation back by applying multiplication.
Recently Updated Pages
Master Class 10 General Knowledge: Engaging Questions & Answers for Success

Master Class 10 Computer Science: Engaging Questions & Answers for Success

Master Class 10 English: Engaging Questions & Answers for Success

Master Class 10 Social Science: Engaging Questions & Answers for Success

Master Class 10 Maths: Engaging Questions & Answers for Success

Master Class 10 Science: Engaging Questions & Answers for Success

Trending doubts
The shortest day of the year in India

Why is there a time difference of about 5 hours between class 10 social science CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

What is the median of the first 10 natural numbers class 10 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

State and prove converse of BPT Basic Proportionality class 10 maths CBSE

