How do you factor $2{{x}^{4}}+5{{x}^{2}}+3$ ?
Answer
616.2k+ views
Hint: Since the given expression has an odd number of terms, we factorize this given polynomial by splitting the middle term $5{{x}^{2}}$. To split the middle term, we use the sum and product form. Now take-out regular terms from the initial two terms and last two terms. At that point keep in touch with them together as the result the product of sums form. Now address them as the components of the given expression.
Complete step by step solution:
The given polynomial which must be factorized is $2{{x}^{4}}+5{{x}^{2}}+3$
To factorize this polynomial, we must first split the middle term.
For this, we use the product sum form.
If the polynomial’s general form is $a{{x}^{2}}+bx+c=0$ or $a{{x}^{4}}+b{{x}^{2}}+c=0$
Either way, we shall take the middle term and the end terms only.
So, the degree doesn’t matter only that the number terms must be odd.
Then to split the middle term, we use the product and sum formula.
Which is,
The product of the middle terms after splitting must be $a\times c$
And the sum of the middle terms must be $b$
Here $a=2;b=5;c=3$
The product of the terms is $2\times 3=6$
The sum of the terms is $5$
Therefore, the terms can be $2{{x}^{2}},3{{x}^{2}}$
On substituting back, we get,
$\Rightarrow 2{{x}^{4}}+2{{x}^{2}}+3{{x}^{2}}+3$
The polynomial is of degree $4$
Now, firstly let us take the common terms out of the first two terms.
$\Rightarrow 2{{x}^{2}}\left( {{x}^{2}}+1 \right)+3{{x}^{2}}+3$
Now secondly take the common terms out of the last two terms.
$\Rightarrow 2{{x}^{2}}\left( {{x}^{2}}+1 \right)+3\left( {{x}^{2}}+1 \right)$
Now on writing it in the form of the product of sums, also known as factoring,
$\Rightarrow \left( 2{{x}^{2}}+3 \right)\left( {{x}^{2}}+1 \right)$
Now writing all the factors together we get,
$\Rightarrow \left( {{x}^{2}}+1 \right)\left( 2{{x}^{2}}+3 \right)$
Hence the factors for the polynomial $2{{x}^{4}}+5{{x}^{2}}+3$ are $\left( {{x}^{2}}+1 \right)\left( 2{{x}^{2}}+3 \right)$.
Note: The given expression is a polynomial of degree four. Hence after factoring we must either get factors in which the equations are or degree $3,1\;$ or $2,2\;$ or else $2,1,1\;$ . To check if our solution is correct, we can always multiply back our factors and check if we get the question back as the solution.
Complete step by step solution:
The given polynomial which must be factorized is $2{{x}^{4}}+5{{x}^{2}}+3$
To factorize this polynomial, we must first split the middle term.
For this, we use the product sum form.
If the polynomial’s general form is $a{{x}^{2}}+bx+c=0$ or $a{{x}^{4}}+b{{x}^{2}}+c=0$
Either way, we shall take the middle term and the end terms only.
So, the degree doesn’t matter only that the number terms must be odd.
Then to split the middle term, we use the product and sum formula.
Which is,
The product of the middle terms after splitting must be $a\times c$
And the sum of the middle terms must be $b$
Here $a=2;b=5;c=3$
The product of the terms is $2\times 3=6$
The sum of the terms is $5$
Therefore, the terms can be $2{{x}^{2}},3{{x}^{2}}$
On substituting back, we get,
$\Rightarrow 2{{x}^{4}}+2{{x}^{2}}+3{{x}^{2}}+3$
The polynomial is of degree $4$
Now, firstly let us take the common terms out of the first two terms.
$\Rightarrow 2{{x}^{2}}\left( {{x}^{2}}+1 \right)+3{{x}^{2}}+3$
Now secondly take the common terms out of the last two terms.
$\Rightarrow 2{{x}^{2}}\left( {{x}^{2}}+1 \right)+3\left( {{x}^{2}}+1 \right)$
Now on writing it in the form of the product of sums, also known as factoring,
$\Rightarrow \left( 2{{x}^{2}}+3 \right)\left( {{x}^{2}}+1 \right)$
Now writing all the factors together we get,
$\Rightarrow \left( {{x}^{2}}+1 \right)\left( 2{{x}^{2}}+3 \right)$
Hence the factors for the polynomial $2{{x}^{4}}+5{{x}^{2}}+3$ are $\left( {{x}^{2}}+1 \right)\left( 2{{x}^{2}}+3 \right)$.
Note: The given expression is a polynomial of degree four. Hence after factoring we must either get factors in which the equations are or degree $3,1\;$ or $2,2\;$ or else $2,1,1\;$ . To check if our solution is correct, we can always multiply back our factors and check if we get the question back as the solution.
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