Give two or more solutions to x< 29. Show why each value is a solution.
1 answer:
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Answer:
-x^3+5x^2-8x+1, which is choice A
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Work Shown:
f(x) = x^3 - x^2 - 3
f(x) = (x)^3 - (x)^2 - 3
f(2-x) = (2-x)^3 - (2-x)^2 - 3 ................ see note 1 (below)
f(2-x) = (2-x)(2-x)^2 - (2-x)^2 - 3 ........... see note 2
f(2-x) = (2-x)(4-4x+x^2) - (4-4x+x^2) - 3 ..... see note 3
f(2-x) = -x^3+6x^2-12x+8 - (4-4x+x^2) - 3 ..... see note 4
f(2-x) = -x^3+6x^2-12x+8 - 4+4x-x^2 - 3 ....... see note 5
f(2-x) = -x^3+5x^2-8x+1
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note1: I replaced every copy of x with 2-x. Be careful to use parenthesis so that you go from x^3 to (2-x)^3, same for the x^2 term as well.
note2: The (2-x)^3 is like y^3 with y = 2-x. We can break up y^3 into y*y^2, so that means (2-x)^3 = (2-x)(2-x)^2
note3: (2-x)^2 expands out into 4-4x+x^2 as shown in figure 1 (attached image below). I used the box method for this and for note 4 as well. Each inner box or cell is the result of multiplying the outside terms. Example: in row1, column1 we have 2 times 2 = 4. You could use the FOIL rule or distribution property, but the box method is ideal so you don't lose track of terms.
note4: (2-x)(4-4x+x^2) turns into -x^3+6x^2-12x+8 when expanding everything out. See figure 2 (attached image below). Same story as note 3, but it's a bit more complicated.
note5: distribute the negative through to ALL the terms inside the parenthesis of (4-4x+x^2) to end up with -4+4x-x^2
