Thanks for the answer Reiny, could you explain the way you solved the factor of 21y^2+38+5
Certainly! To factor the expression 21y^2 + 38y + 5, we need to find two binomials that multiply together to give us this quadratic equation.
Step 1: Look for a common factor, if any. In this case, we don't have any common factors other than 1.
Step 2: Check if the quadratic equation can be factored using the AC method. In this case, the AC value is (21)(5) = 105. We need to find two numbers whose product is 105 and sum is 38.
Step 3: To factor 105, we can consider pairs of factors: (1, 105), (3, 35), (5, 21), (7, 15). Notice that the pair (7, 15) gives us a sum of 22, which is close to 38. But it's not a perfect match.
Step 4: Now, split the middle term (38y) using the pair (7, 15). This means we rewrite the middle term using the sum of these two numbers: 38 = 7 + 15.
Step 5: Now we can rewrite the quadratic equation as: 21y^2 + 7y + 15y + 5
Step 6: Group the terms: (21y^2 + 7y) + (15y + 5)
Step 7: Factor out the greatest common factor from each group: 7y(3y + 1) + 5(3y + 1)
Step 8: Combine the common factors: (7y + 5)(3y + 1)
So, the factors of 21y^2 + 38y + 5 are (7y + 5)(3y + 1).