Chapter 8.7, Problem 2AGP

To determine

To factor: The given polynomial.

Answer to Problem 2AGP

The factored polynomial is $9y^2+24y+16=\left(3y+4\right)\left(3y+4\right)$ .

Explanation of Solution

Given information:

The given polynomial is $9y^2+24y+16$ .

Concept used:

A polynomial of the form $x^2+(a+b)x+ab$ can be factored as $\left(x+a\right)\left(x+b\right)$ , or by British method and 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 $E=9y^2+24y+16$ .

  $E=9y^2+24y+16$

We find two factors of $9\times 16$ such that their sum is 24 as follows:

  $9\times 16=3\times 3\times 4\times 4=\left(3\times 4\right)\times \left(3\times 4\right)=12\times 12$

  $24=12+12$

Now E may be written as $E=9y^2+\left(12+12\right)y+16$

  $\Rightarrow E=9y^2+12y+12y+16$

  $\Rightarrow E=3y\left(3y+4\right)+4\left(3y+4\right)$

  $\Rightarrow E=\left(3y+4\right)\left(3y+4\right)$

Thus, the factored polynomial is $E=9y^2+24y+16=\left(3y+4\right)\left(3y+4\right)$ .

Conclusion:

The factored polynomial is $9y^2+24y+16=\left(3y+4\right)\left(3y+4\right)$ .