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What is x squared minus10 over x squared minus 4 plus 3x over x squared minus 4?
For any x ≠ 0, x2 -10/x2 - 4 +3x/x2 - 4 LCD = x2, multiply each term by their missing element of LCD = (x4 + 10 +3x - 8x2)/x = (x4 - 8x2 + 3x + 10)/x2
What is 2x to the third power plus 8x to the second power plus 3x plus 12 when factoring each polynomial by grouping?
2x3 + 8x2 + 3x + 12 = (2x3 + 8x2) + (3x + 12) = 2x2(x + 4) + 3(x + 4) = (x + 4)(2x2 + 3) Since you have asked this question I am assuming that you are not yet at a level where you might want (2x2 + 3) factorised into its imaginary factors.
X plus 4 x plus 2 simplfied?
x4 +x2 =x2 (x2+1)
What two facts can you double to find 8 x 4?
8x2 8x2
What is the missing factors of 4x7 equals blank x4?
if 4 * 7 = x * 4, then x = (4 * 7) / 4 = 7
What is x squared minus10 over x squared minus 4 plus 3x over x squared minus 4?
For any x ≠ 0, x2 -10/x2 - 4 +3x/x2 - 4 LCD = x2, multiply each term by their missing element of LCD = (x4 + 10 +3x - 8x2)/x = (x4 - 8x2 + 3x + 10)/x2
What answer is equal to the sum in the expression below (3x3 plus 4x2 - 2x plus 1) plus (x4 - x3 plus 5x2 plus 2x plus 3)?
Answer this question…A. x4 + 2x3 + 9x2 + 4 B. x4 + 4x3 + 9x2 + 4 C. x4 + 2x3 + 9x2 + 4x + 4 D. x4 + 2x3 + 9x2 - 4x + 4
4 quebed plus 13 --4 plus 14 squared answer?
4 x4 x4=64 + 13=77- - 4=81+(14x14=196)=277
What is 2x to the third power plus 8x to the second power plus 3x plus 12 when factoring each polynomial by grouping?
2x3 + 8x2 + 3x + 12 = (2x3 + 8x2) + (3x + 12) = 2x2(x + 4) + 3(x + 4) = (x + 4)(2x2 + 3) Since you have asked this question I am assuming that you are not yet at a level where you might want (2x2 + 3) factorised into its imaginary factors.
What is the product of 2X-8 and -4X plus 4?
(2x - 8)(-4x + 4) = -8x2 + 40x - 32
What is an example of factor by grouping?
3x3 + 6x2 + x + 2 3x2 (x + 2) + (x + 2) (x + 2) (3x2 + 1) OR 3x3 + x + 6x2 + 2x (3x2 + 1) + 2 (3x2 + 1) (3x2 + 1) (x + 2) Check: (x + 2) (3x2 + 1) = 3x3 + x + 6x2 + 2 = 3x3 + 6x2 + x + 2 x3 - 3x2 - 4x + 12 x2 (x - 3) - 4 (x - 3) (x - 3) (x2 - 4) (x - 3) (x + 2) (x - 2) Check: (x - 3) (x + 2) (x - 2) = (x - 3) (x2 - 4) = x3 - 4x - 3x2 + 12= x3 - 3x2 - 4x + 12=== === x5 - x4 + 8x3 - 8x2 + 16x - 16 x4 (x - 1) + 8x2 (x - 1) + 16 (x - 1) (x - 1) (x4 + 8x2 + 16) (x - 1) (x2 + 4)(x2 + 4) (x - 1) (x2 + 4)2 Real Solution (x - 1) (x + 2i) (x - 2i) (x + 2i) (x - 2i) (x - 1) (x + 2i)2 (x - 2i)2 Check: (x - 1) (x + 2i)2 (x - 2i)2 = (x - 1) (x + 2i) (x - 2i) (x + 2i) (x - 2i) = (x - 1) (x2 + 2xi - 2xi - 4i2) (x2 + 2xi - 2xi - 4i2) = (x - 1) (x2 - 4(-1)) (x2 - 4(-1)) = (x - 1) (x2 + 4) (x2 + 4) = (x - 1) (x4 + 4x2 + 4x2 +16) = (x - 1) (x4 + 8x2 + 16) = x5 + 8x3 +16x - x4 - 8x2 - 16 = x5 - x4 + 8x3 - 8x2 + 16x - 16
X4-2x3 plus 2x2 plus x plus 4 divided by x2 plus x plus 1 solutions?
(x4 - 2x3 + 2x2 + x + 4) / (x2 + x + 1)You can work this out using long division:x2 - 3x + 4___________________________x2 + x + 1 ) x4 - 2x3 + 2x2 + x + 4x4 + x3 + x2-3x3 + x2 + x-3x3 - 3x2 - 3x4x2 + 4x + 44x2 + 4x + 40R∴ x4 - 2x3 + 2x2 + x + 4 = (x2 + x + 1)(x2 - 3x + 4)
How do you factor x4-x2 plus 2?
x^4 - x^2 plus 2 has no real rational roots.x^4 - x^2 minus 2 factors to (x^2 + 1)(x^2 - 2)
If 4 plus x4 3 plus yy then x4 3y?
The browser used by this site for posting questions is rubbish. We have complained about it but the powers that be are not doing anything about it. The consequence is that most mathematical symbols are not accepted and so all that I can see is of your question is:"If 4 plus x4 3 plus yy then x4 3y?"It is difficult to be sure, but making an informed guess as to what might be missing, I would say that the answer is NO. There is no solution for y.
What is the discriminant of this equation 8x2 plus 5x plus 6 equals 0?
It is: 25-4(8*6) = -167
X plus 4 x plus 2 simplfied?
x4 +x2 =x2 (x2+1)
What is x4 plus x4?
Ah, what a delightful question! When you have x4 plus x4, you can simply add the two terms together. It's like having 4 happy little birds on one branch and then adding 4 more joyful birds to join them, giving you a total of 8 lovely birds sitting together.
