I am trying to figure out the zeroes of 2x^2-7x-30. I did this graphically in a way that matched up with the answers in my book, however, I couldn't get it by distributing. I tried (2x+-10)(x+3) but that didn't get -7x in the middle as logic told me it should. How would you then do this?

2x^2-7x-30 = (2x +5)(x - 6)

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To find the zeroes of the quadratic equation 2x^2 - 7x - 30, you need to factorize the expression correctly. Let's go through the process step by step:

Step 1: Write down the original equation: 2x^2 - 7x - 30 = 0.

Step 2: Look for two numbers whose product is equal to the product of the coefficient of x^2 (2) and the constant term (-30), and whose sum is equal to the coefficient of x (-7).

In this case, we need to find two numbers whose product is -60 and whose sum is -7. By trial and error or using a systematic approach, you can find that -10 and 3 satisfy these conditions.

Step 3: Rewrite the middle term (-7x) as the sum of the two numbers you found in step 2. So, replace -7x with -10x + 3x.

2x^2 - 10x + 3x - 30 = 0

Step 4: Group the first two terms and the last two terms separately:

(2x^2 - 10x) + (3x - 30) = 0

Step 5: Factor out the greatest common factor (GCF) from each grouping:

2x(x - 5) + 3(x - 10) = 0

Step 6: Now, you can see that both groupings have a common binomial factor, (x - 5):

(2x + 3)(x - 5) = 0

Step 7: To find the zeroes, set each factor equal to zero and solve for x:

2x + 3 = 0 or x - 5 = 0

Solving the first equation, you get x = -3/2.

Solving the second equation, you get x = 5.

Therefore, the zeroes of the quadratic equation 2x^2 - 7x - 30 are x = -3/2 and x = 5.

It seems like the mistake you made was in step 3, where you distributed the -10 incorrectly. It should have been -10x instead of -10. By following the correct steps of factorization, you will be able to arrive at the correct answer.