8. Factor 5t² - t -18

9. Factor 25x² - 9

10. Factor 9t² + 12t + 4

See https://www.jiskha.com/display.cgi?id=1522693893#1522693893.1522695283

Factoring takes some practice.

(5t-?)(t+?)
or (5x+?)(t-?) so just play around and you will finally wind up with
10-9=1, so
(5t+9)(t-2) = 5t^2+9t-10t-18 = 5t^2-t-18

also, recall that (a-b)(a+b) = a^2-b^2
25x^2-9 = (5x)^2 - 3^2

10. remember 3*2 = 6

6+6 = 12

To factor a quadratic expression, we are looking for two binomials that multiply together to give us the original expression.

Let's start by factoring each of these expressions one by one:

8. Factor 5t² - t -18:

To begin factoring, we need to break down -t into two terms whose coefficients multiply to give us 5*(-18) = -90, and whose sum is -1 (the coefficient of -t). After some trial and error, we find that -10 and 9 satisfy these conditions. So we can rewrite the expression as:

5t² - t - 18 = 5t² - 10t + 9t - 18

Now we group the terms:

(5t² - 10t) + (9t - 18)

Next, we factor out the greatest common factor from each group:

5t(t - 2) + 9(t - 2)

Notice that we now have a common factor of (t - 2). We can factor this out:

(t - 2)(5t + 9)

Therefore, 5t² - t - 18 factors to (t - 2)(5t + 9).

9. Factor 25x² - 9:

We recognize this expression as the difference of squares: 25x² = (5x)² and 9 = 3².

Using the difference of squares formula, we can write this expression as:

(5x)² - 3² = (5x - 3)(5x + 3)

Therefore, 25x² - 9 factors to (5x - 3)(5x + 3).

10. Factor 9t² + 12t + 4:

To begin factoring, we need to find two numbers whose sum is 12 and whose product is 9 * 4 = 36.

After some trial and error, we find that 6 and 6 satisfy these conditions. So we can rewrite the expression as:

9t² + 6t + 6t + 4

Now we group the terms:

(9t² + 6t) + (6t + 4)

Next, we factor out the greatest common factor from each group:

3t(3t + 2) + 2(3t + 2)

Notice that we now have a common factor of (3t + 2). We can factor this out:

(3t + 2)(3t + 2)

Therefore, 9t² + 12t + 4 factors to (3t + 2)(3t + 2), or simply (3t + 2)².