
How do you factor ${x^4} - 16$?
Answer
540.9k+ views
Hint: Factorization of any polynomials can be written as the product of its factors having the degree less than or equal to the original polynomial. Here we will use the identity for the difference of two squares twice and will simplify for the resultant required solution.
Complete step-by-step solution:
Square is the number multiplied itself and cube it the number multiplied thrice. Square is the product of same number twice such as ${n^2} = n \times n$ for Example square of $2$ is ${2^2} = 2 \times 2$ simplified form of squared number is ${2^2} = 2 \times 2 = 4$ .
Take the given expression and write the expression in terms of the difference of two squares.
$\Rightarrow {x^4} - 16 = {({x^2})^2} - {(4)^2}$
Now using the difference of squares identity: ${a^2} - {b^2} = (a - b)(a + b)$
$\Rightarrow {x^4} - 16 = ({x^2} - 4)({x^2} + 4)$
Again, using the identity for the difference of two squares in one part of the right hand side of the equation.
$\Rightarrow {x^4} - 16 = [{(x)^2} - {(2)^2}]({x^2} + 4)$
Simplifying the values in the above equation.
$\Rightarrow {x^4} - 16 = ({x^2} + 4)(x + 2)(x - 2)$
This is the required solution.
Additional Information: Know the difference between the squares and square root and apply the concepts accordingly. Square is the number multiplied twice and square-root is denoted by $\sqrt {{n^2}} = \sqrt {n \times n} $ For Example: $\sqrt {{2^2}} = \sqrt 4 = 2$
Note: Always remember the different identities to factorize the polynomials. Always try to convert the polynomials in the form of squares and cubes and then apply its formulas. Constants are the terms with fixed value such as the numbers it can be positive or negative whereas the variables are terms which are denoted by small alphabets such as x, y, z, a, b, etc. Be careful while moving any term from one side to another.
Complete step-by-step solution:
Square is the number multiplied itself and cube it the number multiplied thrice. Square is the product of same number twice such as ${n^2} = n \times n$ for Example square of $2$ is ${2^2} = 2 \times 2$ simplified form of squared number is ${2^2} = 2 \times 2 = 4$ .
Take the given expression and write the expression in terms of the difference of two squares.
$\Rightarrow {x^4} - 16 = {({x^2})^2} - {(4)^2}$
Now using the difference of squares identity: ${a^2} - {b^2} = (a - b)(a + b)$
$\Rightarrow {x^4} - 16 = ({x^2} - 4)({x^2} + 4)$
Again, using the identity for the difference of two squares in one part of the right hand side of the equation.
$\Rightarrow {x^4} - 16 = [{(x)^2} - {(2)^2}]({x^2} + 4)$
Simplifying the values in the above equation.
$\Rightarrow {x^4} - 16 = ({x^2} + 4)(x + 2)(x - 2)$
This is the required solution.
Additional Information: Know the difference between the squares and square root and apply the concepts accordingly. Square is the number multiplied twice and square-root is denoted by $\sqrt {{n^2}} = \sqrt {n \times n} $ For Example: $\sqrt {{2^2}} = \sqrt 4 = 2$
Note: Always remember the different identities to factorize the polynomials. Always try to convert the polynomials in the form of squares and cubes and then apply its formulas. Constants are the terms with fixed value such as the numbers it can be positive or negative whereas the variables are terms which are denoted by small alphabets such as x, y, z, a, b, etc. Be careful while moving any term from one side to another.
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