Solve the following equation by factoring:
6(p^2-1)=5p
To solve the equation by factoring, we need to rearrange it so that one side is equal to zero.
Let's start by distributing 6 on the left side of the equation:
6p^2 - 6 = 5p
Next, bring all terms to one side of the equation:
6p^2 - 5p - 6 = 0
Now, we have a quadratic equation in the form of ax^2 + bx + c = 0, where a = 6, b = -5, and c = -6.
To factor the quadratic equation, we need to find two numbers that multiply to give ac (6 * -6 = -36) and add up to give b (-5). In this case, the numbers are -9 and 4.
So, we rewrite -5p as -9p + 4p:
6p^2 - 9p + 4p - 6 = 0
Then, we group the terms and factor by grouping:
(6p^2 - 9p) + (4p - 6) = 0
Take out the greatest common factor from each group:
3p(2p - 3) + 2(2p - 3) = 0
Now, notice that we have (2p - 3) appearing in both terms. We can factor it out:
(2p - 3)(3p + 2) = 0
To find the solutions, we set each factor equal to zero:
2p - 3 = 0 or 3p + 2 = 0
Solving for p in the first equation:
2p = 3
p = 3/2 or p = 1.5
Solving for p in the second equation:
3p = -2
p = -2/3 or p ≈ -0.67
Therefore, the solutions to the equation 6(p^2-1) = 5p are p = 3/2, p = 1.5, p = -2/3, and p ≈ -0.67.
6(p^2-1) = 5p
6p^2 - 5p - 6 = 0
However, I cannot find the factors. Is there a typo?
Sorry.