Chapter 8.6, Problem 1BGP

To determine

To factor: The polynomial $9+10t+t^2$

Answer to Problem 1BGP

  $\left(t+9\right)\left(t+1\right)$

Explanation of Solution

Given:

  $9+10t+t^2$

Calculation:

The given polynomial is $9+10t+t^2$

Write it in standard form

  $t^2+10t+9$

Rewrite the middle term as $10t=9t+t$ . So, the polynomial becomes

  $t^2+9t+t+9$

Make group as shown below

  $\left(t^2+9t\right)+\left(t+9\right)$

Factor out the GCF from each of the group

  $t\left(t+9\right)+1\left(t+9\right)$

Finally, factored out the common factor

  $\left(t+9\right)\left(t+1\right)$

Therefore, the factored form of the given polynomial is $\left(t+9\right)\left(t+1\right)$