Answer:
4t(t+3), 5k²(2k-3), 4x²(2x-5), t=0 and t=1/3, y=0 and y=-2, n=0 and n=-7/4
Step-by-step explanation:
To factor these, we can factor out a constant and/or variable. For number 12, these terms have 4t in common (you can divide any term in the expression by 4t and have no remainder), so we can pull that out. This makes 4t(t+3) because if you distributed, you would end up with the same answer. The answer for 12 is 4t(t+3).
For 13, we do the same thing, except with 5k². When we pull this out of the equation, we end up with 5k²(2k-3), which is the factored expression.
For 14, we can take out 4x². This becomes 4x²(2x-5), the factored expression.
For 15, we actually have to solve, so we can take out a t. This is t(3t-1)=0. We can set both terms (one outside the parentheses and the other one inside the parentheses) equal to zero to solve. t=0, 3t-1=0. To solve the second equation for t, you use inverse operations and end up with t=1/3. The answers are 0 and 1/3.
For 16, we have to solve as well, so take out a 5y. 5y(y+2)=0. Solve for y again. 5y=0, y+2=0. y=0, y=-2 are the answers.
For 17, take out a 3n. 3n(7+4n)=0. Solve for n. 3n=0, 7+4n=0. n=0, n=-7/4 are the answers.
