> #mathematics , #equations , #algebra > How to solve 6(a²+3a)-168?by factorising? ?
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more than 2 weeks ago

How to solve 6(a²+3a)-168?by factorising? ?

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more than 2 weeks ago
To solve the expression 6(a²+3a)-168, you can follow these steps:

1. Start by simplifying the expression inside the parentheses, which is a²+3a. This can be done by factoring out the greatest common factor, which is 'a'.

a² + 3a = a(a + 3)

2. Now substitute this simplified expression back into the original expression:

6(a²+3a)-168 = 6(a(a + 3)) - 168

3. Distribute the 6 to both terms inside the parentheses:

6(a(a + 3)) - 168 = 6a(a) + 6a(3) - 168

Simplify further:

= 6a² + 18a - 168

4. Now, the expression is in the form of a quadratic equation, which can be solved by factoring, completing the square, or using the quadratic formula. However, in this case, the expression cannot be easily factored.

5. To find the values of 'a' that satisfy the equation, you can set the expression equal to zero and solve for 'a'.

6a² + 18a - 168 = 0

6. You can simplify the equation by dividing all terms by 6:

a² + 3a - 28 = 0

7. Now, you can factor the quadratic equation:

(a + 7)(a - 4) = 0

8. Set each factor equal to zero and solve for 'a':

a + 7 = 0 or a - 4 = 0

a = -7 or a = 4

Therefore, the solutions to the equation 6(a²+3a)-168 = 0 are a = -7 and a = 4.
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