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Question 698324: Completely factor the polynomial 3x*3+12x*2+3x+12
Factor the trinomial completely x*2+15x+29
Factor the trinomial completely x*2-12x+35
Factor the trinomial. X*2-4x-5
Factor the trinomial completely. Y*3-15y*2+35y
Factor the trinomial. 4b*2-4b-15
Factor the trinomial completely. 4x*3+10*2-50x
Factor x*3-512
Solve the equation (x-3)(2x+1)=0
Solve and check. 2x*2-7=0
Found 2 solutions by lynnlo, Alan3354:
Answer by lynnlo(4176)   (Show Source):
You can put this solution on YOUR website! (1)12(3x+1)
(2)17x+29
(3)-5(2x-7)
(4)-2x-5
(5)8y
(6)4b-15
(7)-2(19x-10)
(8)3x-512
(9)x=3,-1/2
(10)2x*2-7=0======worked out this problem
4x-7=0
4*x-7=0
0=0
x=1.75
Answer by Alan3354(69443)  (Show Source):
You can put this solution on YOUR website! Completely factor the polynomial 3x*3+12x*2+3x+12
Factor out a 3
Use ^, Shift 6, for exponents. * is used for multiply.
The other tutor saw * as multiply.
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= 3*(x^3 + 4x^2 + x + 4)
Try the factors of 4, = 1, 2 & 4. Note that all terms are positive, so no negative factors.
f(x) = x^3 + 4x^2 + x + 4
Try x+1. If it's a factor, then f(-1) will be zero.
f(-1) = -1 + 4 - 1 + 4 <> 0
x+1 doesn't work.
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Try x+2
f(-2) = -8 + 16 - 2 + 4 also <> 0
Try x+4
f(-4) = -64 + 64 - 4 + 4 which is zero.
Divide the polynomial by (x+4)
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= 3(x+4)(x^2+1)
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Factor the trinomial completely x*2+15x+29
That can't be factored.
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Factor the trinomial completely x*2-12x+35
Find 2 factors of 35, both negative, that add to -12.
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Factor the trinomial. X*2-4x-5
Easy one.
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Factor the trinomial completely. Y*3-15y*2+35y
Factor out a y, then look for 2 factors of 35 (both -) that add to -15.
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Factor the trinomial. 4b*2-4b-15
This one has more possibilities, both end terms have more than 1 pair of factors.
It could be (b + p)*(4b + q) or (2b + p)*(2b + q), and the signs of p & q are opposite.
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Factor the trinomial completely. 4x*3+10*2-50x
Step 1, factor out 2x.
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Factor x*3-512
A difference of 2 cubes.
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Solve the equation (x-3)(2x+1)=0
If the product is zero, either
x-3 = 0
or
2x+1 = 0
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Solve and check. 2x*2-7=0
Plus and minus.
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