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# Factorise $xy\left( {{z}^{2}}+1 \right)+z\left( {{x}^{2}}+{{y}^{2}} \right).$(a) $\left( xy+z \right)\left( yz+x \right)$(b) $\left( zx+y \right)\left( xy+z \right)$(c) $\left( yz+x \right)\left( zx+y \right)$(d) $\left( {{x}^{2}}+{{y}^{2}} \right)\left( z+x \right)$

Last updated date: 22nd Feb 2024
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Hint: In the above question, we are having a polynomial equation. We have to factorise this polynomial equation. Now, we know that factorization is the process in which we decompose or make factors of the polynomial equation or entity. These factors can be any real number, linear or polynomial equations which are multiplied together to form a single polynomial equation. Now, to factorise this polynomial equation, we will deduce the common factors present in the different part of the polynomial equation.

Complete step-by-step solution:
In the above question, we are given a polynomial equation $xy\left( {{z}^{2}}+1 \right)+z\left( {{x}^{2}}+{{y}^{2}} \right).$ to factorise.
Now, to factorise this polynomial equation, we will simplify the equation.
Now, on simplifying,
We get,
$\Rightarrow xy\left( {{z}^{2}}+1 \right)+z\left( {{x}^{2}}+{{y}^{2}} \right)$
$\Rightarrow xy{{z}^{2}}+xy+z{{x}^{2}}+z{{y}^{2}}$
Now, gathering the common factors in the polynomial equation together,
We get,
$\Rightarrow xz\left( yz+x \right)+y\left( yz+x \right)$
Now, taking $\left( yz+x \right)$ common in the polynomial equation,
We get,
$\Rightarrow \left( xz+y \right)\left( yz+x \right)$
Now, the factored form of the polynomial equation $xy\left( {{z}^{2}}+1 \right)+z\left( {{x}^{2}}+{{y}^{2}} \right)$ is $\left( xz+y \right)\left( yz+x \right)$
Now, checking all the options one by one for the correct option,
Option A: this option states that the factored form of the polynomial equation $xy\left( {{z}^{2}}+1 \right)+z\left( {{x}^{2}}+{{y}^{2}} \right)$ is $\left( xy+z \right)\left( yz+x \right)$ . But this is incorrect because we have proved that the factored form is $\left( xz+y \right)\left( yz+x \right)$. Hence, this option is incorrect.
Option B: this option states that the factored form of the polynomial equation $xy\left( {{z}^{2}}+1 \right)+z\left( {{x}^{2}}+{{y}^{2}} \right)$ is $\left( zx+y \right)\left( xy+z \right)$ . But this is incorrect because we have proved that the factored form is $\left( xz+y \right)\left( yz+x \right)$. Hence, this option is incorrect.
Option C: this option states that the factored form of the polynomial equation $xy\left( {{z}^{2}}+1 \right)+z\left( {{x}^{2}}+{{y}^{2}} \right)$ is $\left( yz+x \right)\left( zx+y \right)$ . As we have proved above that this is the correct factored form of the polynomial equation given in the question. Hence, this is the correct option.
Option D: this option states that the factored form of the polynomial equation $xy\left( {{z}^{2}}+1 \right)+z\left( {{x}^{2}}+{{y}^{2}} \right)$ is $\left( {{x}^{2}}+{{y}^{2}} \right)\left( z+x \right)$ . But this is incorrect because we have proved that the factored form is $\left( xz+y \right)\left( yz+x \right)$. Hence, this option is incorrect.

Hence, the correct option is $C$.

Note: Factorization or factoring consist of writing a number or another mathematical object as a product of several factors, usually smaller or simpler objects of the same kind. For example, $3\times 5$ is a factorization of the integer $15$, and is a factorization of the polynomial ${{x}^{2}}-4.$
We know that we use factorization, to simplify the equation. This method is very useful to find factors and to get a complex equation simplified. Here we have factored the equation given in the question by taking common factors present in the polynomial equation and then written factors in one bracket and common term in another bracket.