
How do you factor \[{x^4} + 2{x^2} + 1\] ?
Answer
558.9k+ views
Hint: The given equation is a biquadratic equation. Here, in this question, we find the factors by different methods firstly by using the method of factorisation and also by using the sum product rule for the equation and hence we find the factors for the equation.
Complete step-by-step answer:
Factorization means the process of creating a list of factors otherwise in mathematics, factorization or factoring is the decomposition of an object into a product of other objects, or factors, which when multiplied together give the original.
The general form of an equation is \[a{x^2} + bx + c\] when the coefficient of \[{x^2}\] is unity. Every quadratic equation can be expressed as \[{x^2} + bx + c = \left( {x + d} \right)\left( {x + e} \right)\] . Here, b is the sum of d and e and c is the product of d and e.
Consider the given expression \[{x^4} + 2{x^2} + 1\] .
Temporarily Rewrite \[{x^4} = {\left( {{x^2}} \right)^2} = {u^2}\] , where \[u = {x^2}\]
The given expression can be written as
\[ \Rightarrow {u^2} + 2u + 1\]
consider the factors of the product ac which sum to b, Product of factors of 1 are \[1 = 1 \times 1\] , \[ - 1 \times - 1\] Here, the possible pair of factor gives a summation of 2 is \[\left( {1,1} \right)\] .
Now, Break the middle term as the summation of two numbers such that its product is equal to 1. Calculated above such two numbers are 1 and 1.
\[ \Rightarrow {u^2} + 1.u + 1.u + 1\]
\[ \Rightarrow {u^2} + u + u + 1\]
Making pairs of terms in the above expression
\[ \Rightarrow \left( {\,{u^2} + u} \right) + \left( {u + 1} \right)\]
Take out greatest common divisor GCD from the both pairs, then
\[ \Rightarrow u\left( {\,u + 1} \right) + 1\left( {u + 1} \right)\]
Take \[\left( {u + 1} \right)\] common
\[ \Rightarrow \left( {u + 1} \right)\left( {u + 1} \right)\]
\[ \Rightarrow {\left( {u + 1} \right)^2}\]
Again reverse replace \[u = {x^2}\]
\[ \Rightarrow {\left( {{x^2} + 1} \right)^2}\]
Hence, the factors of the expression \[{x^4} + 2{x^2} + 1\] is \[\,{\left( {{x^2} + 1} \right)^2}\] .
So, the correct answer is “ \[\,{\left( {{x^2} + 1} \right)^2}\] ”.
Note: We have converted the equation to a quadratic equation by substituting the square term of x as u. This problem can be solved by using the sum product rule. This defines as for the general quadratic equation \[a{x^2} + bx + c\] , the product of \[a{x^2}\] and c is equal to the sum of bx of the equation. Hence, we obtain the factors.
Complete step-by-step answer:
Factorization means the process of creating a list of factors otherwise in mathematics, factorization or factoring is the decomposition of an object into a product of other objects, or factors, which when multiplied together give the original.
The general form of an equation is \[a{x^2} + bx + c\] when the coefficient of \[{x^2}\] is unity. Every quadratic equation can be expressed as \[{x^2} + bx + c = \left( {x + d} \right)\left( {x + e} \right)\] . Here, b is the sum of d and e and c is the product of d and e.
Consider the given expression \[{x^4} + 2{x^2} + 1\] .
Temporarily Rewrite \[{x^4} = {\left( {{x^2}} \right)^2} = {u^2}\] , where \[u = {x^2}\]
The given expression can be written as
\[ \Rightarrow {u^2} + 2u + 1\]
consider the factors of the product ac which sum to b, Product of factors of 1 are \[1 = 1 \times 1\] , \[ - 1 \times - 1\] Here, the possible pair of factor gives a summation of 2 is \[\left( {1,1} \right)\] .
Now, Break the middle term as the summation of two numbers such that its product is equal to 1. Calculated above such two numbers are 1 and 1.
\[ \Rightarrow {u^2} + 1.u + 1.u + 1\]
\[ \Rightarrow {u^2} + u + u + 1\]
Making pairs of terms in the above expression
\[ \Rightarrow \left( {\,{u^2} + u} \right) + \left( {u + 1} \right)\]
Take out greatest common divisor GCD from the both pairs, then
\[ \Rightarrow u\left( {\,u + 1} \right) + 1\left( {u + 1} \right)\]
Take \[\left( {u + 1} \right)\] common
\[ \Rightarrow \left( {u + 1} \right)\left( {u + 1} \right)\]
\[ \Rightarrow {\left( {u + 1} \right)^2}\]
Again reverse replace \[u = {x^2}\]
\[ \Rightarrow {\left( {{x^2} + 1} \right)^2}\]
Hence, the factors of the expression \[{x^4} + 2{x^2} + 1\] is \[\,{\left( {{x^2} + 1} \right)^2}\] .
So, the correct answer is “ \[\,{\left( {{x^2} + 1} \right)^2}\] ”.
Note: We have converted the equation to a quadratic equation by substituting the square term of x as u. This problem can be solved by using the sum product rule. This defines as for the general quadratic equation \[a{x^2} + bx + c\] , the product of \[a{x^2}\] and c is equal to the sum of bx of the equation. Hence, we obtain the factors.
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