Chapter 8.7, Problem 1DGP

To determine

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

Answer to Problem 1DGP

The polynomial $64g^2-h^2$ is a difference of two squares, so it can be factorized. The factored polynomial is ${\left(8g\right)}^2-{\left(h\right)}^2=\left(8g-h\right)\left(8g+h\right)$ .

Explanation of Solution

Given information:

The given polynomial is $64g^2-h^2$ .

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 $P=64g^2-h^2$ .

The polynomial can be written as $P={\left(8g\right)}^2-{\left(h\right)}^2$ , i.e., difference of two squares.

The factors of polynomial P are

  $P={\left(8g\right)}^2-{\left(h\right)}^2$

  $\Rightarrow P=\left(8g-h\right)\left(8g+h\right)$

Thus, the given polynomial $64g^2-h^2$ is a difference of two squares, so it can be factorized.

The factored polynomial is $P={\left(8g\right)}^2-{\left(h\right)}^2=\left(8g-h\right)\left(8g+h\right)$ .

Conclusion:

The polynomial $64g^2-h^2$ is a difference of two squares, so it can be factorized. The factored polynomial is ${\left(8g\right)}^2-{\left(h\right)}^2=\left(8g-h\right)\left(8g+h\right)$ .