Question

which of the following is a factor of ( x^2 - 3x - 28 )?
a ( (x - 28) )
b ( (x + 2) )
c ( (x - 8) )
d ( (x - 7) )
e none of these

Explanation:

Step1: Factor the quadratic

We need to factor \(x^2 - 3x - 28\). We look for two numbers that multiply to \(-28\) and add to \(-3\). The numbers are \(-7\) and \(4\) since \((-7)\times4=-28\) and \(-7 + 4=-3\). So, \(x^2-3x - 28=(x - 7)(x+4)\).

Step2: Check the options

• Option A: \((x - 27)\) is not a factor.
• Option B: \((x + 2)\) is not a factor.
• Option C: \((x - 8)\) is not a factor.
• Option D: \((x + 4)\) is a factor (from the factoring above). Wait, but looking at the options again, maybe there was a typo? Wait, no, let's re - check. Wait, the factored form is \((x - 7)(x + 4)\). So among the given options, if we consider the options, maybe the user made a typo, but let's check again. Wait, the original quadratic is \(x^2-3x - 28\). Let's test each option by plugging into the quadratic formula or by polynomial division. Let's use the factor theorem: if \((x - a)\) is a factor of \(f(x)\), then \(f(a)=0\). Let \(f(x)=x^2-3x - 28\).
• For option A: \(f(27)=27^2-3\times27 - 28=729-81 - 28=620

eq0\).

• For option B: \(f(-2)=(-2)^2-3\times(-2)-28 = 4 + 6-28=-18

eq0\).

• For option C: \(f(8)=8^2-3\times8 - 28=64-24 - 28=12

eq0\).

• For option D: \(f(-4)=(-4)^2-3\times(-4)-28=16 + 12-28=0\). So \((x + 4)\) is a factor. But wait, in the options, if option D is \((x + 4)\), then that's the factor. But maybe the original problem had a different quadratic? Wait, no, the quadratic is \(x^2-3x - 28\). So the correct factor among the options (if option D is \((x + 4)\)) is \((x + 4)\). But let's check the options again. Wait, the user's options: A is \((x - 27)\), B is \((x + 2)\), C is \((x - 8)\), D is \((x + 4)\), E is "None of these". So since we factored \(x^2-3x - 28=(x - 7)(x + 4)\), \((x + 4)\) is a factor, so option D.

Answer:

D. \((x + 4)\)