What are you supposed to get when you factor 3q^2+16q=-5? I got (q+3)(3q+1)
3q^2+16q= -5
3q^2+16q+5 = 0 now look for factors of 15 that add to be 16 (1 &15)
so (3q + 1)(q +1) = 0
so (3q + 1) = 0 and/or (q +1) = 0
so{-1/3, -1} are your possible solutions
How did you get (3q + 1)(q +1) = 0? What happened to the 15?
There is a typo in the previous information.
3q^2 + 16q + 5 = (3q + 1)(q + 5) = 0
3 * 5 = 15
so q could = -1/3 or -5
This should make it clearer. Thanks for asking.
To factor the quadratic expression 3q^2 + 16q = -5, you need to find two binomials that can multiply together to give you the original expression.
Here's how you can determine the factors:
Step 1: Write down the quadratic expression in the general form: ax^2 + bx + c = 0. In this case, a = 3, b = 16, and c = -5. Rewrite the equation as follows: 3q^2 + 16q + 5 = 0.
Step 2: Multiply the coefficient of the quadratic term "a" (3) by the constant term "c" (5). In this case, 3 * 5 = 15.
Step 3: Find two numbers that multiply to give 15 and add up to the coefficient of the linear term "b" (16). In this case, the numbers are 15 and 1 since 15 * 1 = 15 and 15 + 1 = 16.
Step 4: Rewrite the linear term (16q) as the sum of the two numbers found in the previous step. Replace the middle term of the quadratic expression with these two numbers: 3q^2 + 15q + q + 5 = 0.
Step 5: Group the terms and factor by grouping:
(3q^2 + 15q) + (q + 5) = 0.
Step 6: Factor out the greatest common factor (GCF) from each group:
3q(q + 5) + 1(q + 5) = 0.
Step 7: Notice that both groups (q + 5) are the same. Factor out this common binomial:
(q + 5)(3q + 1) = 0.
Therefore, the factored form of the quadratic expression 3q^2 + 16q = -5 is (q + 5)(3q + 1).