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.