How would you solve questions like these I am still very confused I missed a few days of school...

Question

1). 10t2+13t-3. (The 2 is an exponent)

2). W6-4. (The 6 is an exponent)

Answer 1

The second problem is a great example of something that relates to what you have learned in class but isn't the same format as what you have seen.  So what I usually tell students is to think about the topics you have learned and how they might apply to the problem.  You will find that this often lets you figure out the most direct way of solving the problem (in this case, difference of squares).

Answer 2

Hi Maureen,

It is unclear from your question whether you are just being asked to factor these expressions or to solve equations in which they are set equal to 0.  Either way, we need to factor them.  So, I'll show you how to do that and then how to solve the equations if they were set equal to 0.

10t² + 13t - 3 is a quadratic expression.  So, we want to try to factor it as two monomials like (at +/- c) and (bt +/- c) where a and b are factors of 10 and c and d are factors of 3.  We hope to find factors such that (at+/- c)*(bt +/- d) equals the original expression.  The key is to make sure that you end up with +13t when multiplying.  But that is not always possible.  If you try all the possible combinations and none of them work, you will then have to try the quadratic equation.

Let's try some possible factors.  10 can be factored as 1*10 and 2*5 while 3 can be factored as 1*3.

(10t - 3)*(1t + 1) = 10t² + 10t - 3t -3 = 10t² + 7t - 3: the 10t² and -3 are good, but +7t is wrong.

(5t - 3)*(2t + 1) = 10t² + 5t -6t - 3 = 10t² -1t - 3: the 10t² and -3 are good, but -1t is wrong.

(5t - 1)*(2t + 3) = 10t² + 15t - 2t -3 = 10t² + 13t -3: that's right!

Now, if we were asked to solve 10t² + 13t - 3 = 0, we would end up with:

(5t - 1)*(2t + 3) = 0

So, either 5t - 1 or 2t + 3 has to be 0.  If 5t - 1 = 0, we get 5t = 1 and then t = 1/5.  If 2t + 3 = 0, we get 2t = -3 and then t = -3/2.  So, the solutions would be t=1/5 and t=-3/2.

For the second problem, W⁶ - 4, I suggest you think of W⁶ as (W³)².  The expression is then a difference of squares (W³)² - 2² and you should be able to factor this in terms of W³ by using standard methods for factoring x² - a² where a is a constant.  You'll get (W³ + 2)*(W³ - 2).  If solving the equation W⁶ - 4 = 0, you could set these two factors to 0 and solve for W³ = -2 and 2 by taking the cube root (1/3 power) of -2 and +2.

Hope this helps!

Roger