Chapter 8.7, Problem 43PPS

To determine

To factor the given polynomial

Answer to Problem 43PPS

  $(y^2-2)(y^2+2)(y^4+4)(y^6+16).$

Explanation of Solution

Given:

Polynomial: $y^8-256$

Concept used:

Finding common factor in each term and factoring using difference of squares.

Calculation:

Consider the given polynomial,

  $\Rightarrow y^8-256$

Further, $y^8-256$ can be factored as,

  $\begin{array}{l}$ \Rightarrow y^8-256=({(y^6)}^2-{16}^2)=(y^6-16)(y^6+16)$\because a^2-b^2=(a+b)(a-b)\end{array}$

  $(y^6-16)$ can be further factorised as,

  $\begin{array}{l}\Rightarrow (y^6-16)\\ $ \Rightarrow ({(y^4)}^2)-4^2)$\Rightarrow (y^4-4)(y^4+4)\end{array}$

  $(y^4-4)$ can be further factorised as,

  $\begin{array}{l}\Rightarrow (y^4-4)\\ $ \Rightarrow ({(y^2)}^2-2^2)$\Rightarrow (y^2-2)(y^2+2)\end{array}$

Therefore, the polynomial can be factorised as,

  $\Rightarrow (y^2-2)(y^2+2)(y^4+4)(y^6+16)$

Conclusion:

Therefore, the polynomial is factored as $(y^2-2)(y^2+2)(y^4+4)(y^6+16).$