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How do you factor 4t2 - 12?

Updated: 8/20/2019
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8y ago

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4(t^2 - 3)

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Q: How do you factor 4t2 - 12?
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Let L be the line in R3 that consists of all scalar multiples of the vector -2 -1 2T Find the reflection of the vector v equal to 7 2 7T in the line L?

I'm not going to lie, I'm not 100% sure if I'm going to do this correctly. It's been a while since I've done something like this, so you may want to double check my answer. Also, the letter T in your question is really throwing me off, so I'm just going to give you two answers. The first answer will treat T as a variable, the second answer will ignore it completely. Both answers, however, use the following equation for the reflection of a vector about a line:RefL(v) = 2L(v ● L)/(L ● L) - vFor my first answer, I'll use the following vectors for Land v:L = -2i - j + 2Tk, andv = 7i + 2j + 7Tk,where i, j, and k are the unit vectors in the direction of the x, y, and z axes in R3, respectively.Thus,v ● L = -14 - 2 + 14T2 = 14T2 - 16.L ● L = 4 + 1 + 4T2 = 4T2 + 5.Therefore, 2L(v ● L)/(L ● L) =2L(14T2 - 16)/(4T2 + 5) = L(28T2 - 32)/(4T2 + 5) =-8(7T2 - 8)/(4T2 + 5)i - 4(7T2 - 8)/(4T2 + 5)j + 8T(7T2 - 8)/(4T2 + 5)k.Let A = 2L(v ● L)/(L ● L).(A - v)i = [-8(7T2 - 8)/(4T2 + 5) - 7(4T2 + 5)/(4T2 + 5)]i =(-56T2 + 64 - 28T2 - 35)/(4T2 + 5)i = (-84T2 + 29)/(4T2 + 5)i.(A - v)j = [-4(7T2 - 8)/(4T2 + 5) - 2(4T2 + 5)/(4T2 + 5)]j =(-28T2 + 32 - 8T2 - 10)/(4T2 + 5)j = (-36T2 + 22)/(4T2 + 5)j.(A - v)k = [8T(7T2 - 8)/(4T2 + 5) - 7T(4T2 + 5)/(4T2 + 5)]k =(56T3 - 64T - 28T3 - 35T)/(4T2 + 5)k = (28T3 - 99T)/(4T2 + 5)k.Let b = 1/(4T2 + 5), thenRefL(v) = b[(-84T2 + 29)i + (-36T2 + 22)j + (28T3 - 99T)k]That expression for RefL(v) looks pretty ugly, so I'm going to do the problem again, this time without the variable T.L = -2i - j + 2k, andv = 7i + 2j + 7k.v ● L = -14 - 2 + 14 = -2L ● L = 4 + 1 + 4 = 9Therefore, 2L(v ● L)/(L ● L) =2L(-2/9) = L(-4/9) =(8/9)i + (4/9)j - (8/9)k.RefL(v) = 2L(v ● L)/(L ● L) - v = (8/9)i + (4/9)j - (8/9)k - 7i - 2j - 7k =-(55/9)i - (14/9)j - (71/9)k.While this expression does look much nicer, I'm not sure if it's right. So, like I recommended above, please double check my work!


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