Chapter 8.7, Problem 5DGP

To determine

To find: That the given polynomial is a difference of squares, if yes then to factor it.

Answer to Problem 5DGP

The polynomial $16y^4-1$ is a difference of two squares, so it is factorable. The factored polynomial is $16y^4-1=\left(2y-1\right)\left(2y+1\right)\left(4y^2+1\right)$ .

Explanation of Solution

Given information:

The given polynomial is $16y^4-1$ .

Concept used:

The difference of two squares can be factored as $a^2-b^2=\left(a-b\right)\left(a+b\right)$ .

Calculations:

The given polynomial P (say) is $P=16y^4-1$ .

The polynomial can be written as $P={\left(4y^2\right)}^2-{\left(1\right)}^2$ , i.e., difference of two squares.

Since, the given polynomial $16y^4-1$ is a difference of two squares, so it can be factorized.

The factors of polynomial P are

  $P={\left(4y^2\right)}^2-{\left(1\right)}^2$

  $\Rightarrow P=\left(4y^2-1\right)\left(4y^2+1\right)$

  $\Rightarrow P=\left({(2y)}^2-{(1)}^2\right)\left(4y^2+1\right)$

One factor again a perfect square

  $\Rightarrow P=\left(2y-1\right)\left(2y+1\right)\left(4y^2+1\right)$

Thus, the given polynomial $16y^4-1$ is twice a difference of two squares, so it can be factorized.

The factored polynomial is $P=16y^4-1=\left(2y-1\right)\left(2y+1\right)\left(4y^2+1\right)$ .

Conclusion:

The polynomial $16y^4-1$ is a difference of two squares, so it is factorable. The factored polynomial is $16y^4-1=\left(2y-1\right)\left(2y+1\right)\left(4y^2+1\right)$ .