Factorize $x^4+64$

The real number $64$ is added to a variable $x$ raised to the power of $4$ for representing an unknown in a polynomial. In this problem, the given algebraic expression has to be factored.

$x^4+64$

Express the expression in sum of squares

The variable $x$ raised to the power $4$ and the real number $64$ can be written in square form. So, they can be written as square of $x$ squared and $8$ squared.

$=\,\,\,$ $\big(x^2\big)^2$ $+$ $8^2$

Simplify expression as square of a Binomial

The algebraic expression is converted as the sum of squares. The sum of squares is actually part of the expansion of square of either sum or difference of two terms. According to them, the polynomial in square form should have a term additionally and it is $2 \times 8 \times x^2$. So, add it to the expression and subtract the same term from their sum for mathematical acceptance.

$=\,\,\,$ $\big(x^2\big)^2$ $+$ $8^2$ $+$ $2 \times x^2 \times 8$ $-$ $2 \times x^2 \times 8$

$=\,\,\,$ $\big(x^2\big)^2$ $+$ $8^2$ $+$ $2 \times 8 \times x^2$ $-$ $2 \times 8 \times x^2$

According to the square of sum of two terms formula, the first three terms can be simplified as the square of sum of $x$ square and $8$.

$=\,\,\,$ $\big(x^2+8\big)^2$ $-$ $16 \times x^2$

Look at the second term in the expression, the second factor is in square form. So, try to express the real number $16$ in square form for expressing their product in square form.

$=\,\,\,$ $\big(x^2+8\big)^2$ $-$ $4^2 \times x^2$

According to the power of a product rule, the product of squares can be simplified as a square of product of their bases.

$=\,\,\,$ $\big(x^2+8\big)^2$ $-$ $(4 \times x)^2$

$=\,\,\,$ $\big(x^2+8\big)^2$ $-$ $(4x)^2$

Factorise by the difference of squares method

The given algebraic expression $x^4+64$ is successfully converted as $\big(x^2+8\big)^2$ $-$ $(4x)^2$. The simplified expression represents difference of the terms and it can be factored as per the factorization by the difference of two squares.

$=\,\,\,$ $\big(x^2+8+4x\big)\big(x^2+8-4x\big)$

$=\,\,\,$ $\big(x^2+4x+8\big)\big(x^2-4x+8\big)$

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