
How do you factor \[3{{x}^{2}}+7x+2\]?
Answer
445.8k+ views
Hint: Apply the middle term split method to factorize \[3{{x}^{2}}+7x+2\]. Split 7x into two terms in such a way that their sum is 7x and the product is \[6{{x}^{2}}\]. For this process, find the prime factors of 6 and combine them in such a way so that we can get our conditions satisfied. Finally, take the common terms together and write \[3{{x}^{2}}+7x+2\] as a product of two terms given as: - \[\left( x-a \right)\left( x-b \right)\], where ‘a’ and ‘b’ are called zeroes of the polynomial.
Complete step-by-step solution:
Here, we have been asked to factorize the quadratic polynomial \[3{{x}^{2}}+7x+2\].
Let us use the middle term split method for the factorization. It states that we have to split method for the factorization. It states that we have to split the middle term which is 7x into two terms such that their sum is 7x and the product is equal to the product of constant term (2) and \[3{{x}^{2}}\], i.e., \[6{{x}^{2}}\]. To do this, first we need to find all the prime factors of 6. So, let us find.
We know that 6 can be written as: - \[6=2\times 3\] as the product of its primes. Now, we have to group these factors such that our conditions of the middle term split method are satisfied. So, we have,
(i) \[\left( x \right)+\left( 6x \right)=7x\]
(ii) \[\left( x \right)\times \left( 6x \right)=6{{x}^{2}}\]
Hence, both the conditions of the middle term split method are satisfied. So, the quadratic polynomial can be written as: -
\[\begin{align}
& \Rightarrow 3{{x}^{2}}+7x+2=3{{x}^{2}}+x+6x+2 \\
& \Rightarrow 3{{x}^{2}}+7x+2=x\left( 3x+1 \right)+2\left( 3x+1 \right) \\
\end{align}\]
Taking \[\left( 3x+1 \right)\] common in the R.H.S., we get,
\[\Rightarrow 3{{x}^{2}}+7x+2=\left( 3x+1 \right)\left( x+2 \right)\]
Hence, \[\left( 3x+1 \right)\left( x+2 \right)\] is the factored form of the given quadratic polynomial.
Note: One may note that we can use another method for the factorization. Discriminant methods can also be applied to solve the question. What we will do is we will find the solution of the quadratic equation using discriminant method. The values of x obtained will be assumed as x = a and x = b. Finally, we will consider the product \[\left( x-a \right)\left( x-b \right)\] to get the factored form.
Complete step-by-step solution:
Here, we have been asked to factorize the quadratic polynomial \[3{{x}^{2}}+7x+2\].
Let us use the middle term split method for the factorization. It states that we have to split method for the factorization. It states that we have to split the middle term which is 7x into two terms such that their sum is 7x and the product is equal to the product of constant term (2) and \[3{{x}^{2}}\], i.e., \[6{{x}^{2}}\]. To do this, first we need to find all the prime factors of 6. So, let us find.
We know that 6 can be written as: - \[6=2\times 3\] as the product of its primes. Now, we have to group these factors such that our conditions of the middle term split method are satisfied. So, we have,
(i) \[\left( x \right)+\left( 6x \right)=7x\]
(ii) \[\left( x \right)\times \left( 6x \right)=6{{x}^{2}}\]
Hence, both the conditions of the middle term split method are satisfied. So, the quadratic polynomial can be written as: -
\[\begin{align}
& \Rightarrow 3{{x}^{2}}+7x+2=3{{x}^{2}}+x+6x+2 \\
& \Rightarrow 3{{x}^{2}}+7x+2=x\left( 3x+1 \right)+2\left( 3x+1 \right) \\
\end{align}\]
Taking \[\left( 3x+1 \right)\] common in the R.H.S., we get,
\[\Rightarrow 3{{x}^{2}}+7x+2=\left( 3x+1 \right)\left( x+2 \right)\]
Hence, \[\left( 3x+1 \right)\left( x+2 \right)\] is the factored form of the given quadratic polynomial.
Note: One may note that we can use another method for the factorization. Discriminant methods can also be applied to solve the question. What we will do is we will find the solution of the quadratic equation using discriminant method. The values of x obtained will be assumed as x = a and x = b. Finally, we will consider the product \[\left( x-a \right)\left( x-b \right)\] to get the factored form.
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