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Find the smalllest zero in h(x)=4x^2-8x-60

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Answer:

x = -3

Explanation:

Step 1: First, let's find the greatest common factor (GCF) of 4, -8, and -60 and factor it out.

  • The GCF of these three numbers is 4, so we can factor it out by dividing each term by 4:

4/4 = 1, -8/4 = -2, and -60/4 = -15.

Thus, the equation we can now use is 4(x^2 - 2x - 15).

Step 2: We can by find the zeroes by factoring.

  • Currently, 4(x^2 - 2x - 15) is in standard form, whose general form is ax^2 + bx + c (disregard the 4 for the moment).
  • Thus, 1 is our a value, -2 is our b value, and -15 is our c value
  • We want to find two numbers whose product equals the product of a and c and whose sum equals b
  • We see that -5 form of quadratics, we have 4(x - 5)(x + 3) = 0
  • In order to solve for x, we set both terms equal to 0 and solve to find the zeroes of the function:

Step 3: Setting 4(x - 5) equal to 0:

4(x - 5) = 0

x - 5 = 0

x = 5

Setting (x + 3) equal to 0:

x + 3 = 0

x = -3

Thus, the smallest zero is -3

User Chris Clark
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