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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
How do you factor completely x to the fourth power minus 1?
So x4 - 1 is the difference of squares, so x4 - 1 = (x2 - 1)(x2 + 1) = (x + 1)(x - 1)(x2 + 1).
When is a polynomial factored completely?
It is factored completely if it can not be factored any more. Quite simply, when all like terms have been combined, and all common factors taken out. 2x + 10y + 8x - original equation 2x + 8x + 10y - rearrange equation (2x + 8x) + 10y - combine terms 10x + 10y - note common factor 10(x+y) - factored 14xy + 10y + 6 - original equation 2(7xy + 5y + 3) - factor out two. 2(y[7x + 5] + 3) - factor out y. This is factored completely. 10x + 3xy + 6y. It is really hard to make a call on this one. You can factor it either like x(10+3y) + 6y, or 10x + 3y(x + 2). However, polynomials become more complicated that this, however. Try factoring this: x2 - 4x + 4 That isn't so easy to factor because there are no like terms. However, it is easy to note that the third term is positive. It is also easy to note that the second term is negative. So, you know that if it is able to be factored, that the factors will both have a minus sign. Using a little deductive reasoning, you'll find that this factors down to (x-2)(x-2), or (x-2)2. How about this? x4 - 1 Note that despite the fact that there is an x4 in this, it still follows the rules of difference of squares. Thus can be factored (x2 - 1)(x2 + 1) Looking at this, you can see, again, another difference of squares. So, we continue factoring. (x+1)(x-1)(x2 + 1). Now comes probably to the heart of the question. How do we know that x2 + 1 is factored as far as it can go? x2 + 1 can be rewritten as x2 + 0x + 1. We can note in this instance that the second and third terms are both positive. That means for this to work, there must be case where two positive terms added together equal zero. (x + )(x + ). By logical determination, we can conclude that there is no such number that makes this possible. Thus, x2 + 1 is factored as far as it can go, and thus: (x+1)(x-1)(x2 + 1) is also completely factored. Another example: x2 -5x - 14 First of all, we start by making observations. Both second and third terms are negative. That means, when we factor this, each binomial factor will have a different sign. We can start by writing out the ground work. (x + )(x - ). Then we try to come up with divisors of 14. In this case: 1, 2, 7, 14. We need to find two that when subtracted equal 5 (different signs equal subtraction when combining like terms). In this case, 7 and 2. Note, for the second term, 5, to be negative, the larger of the two numbers must be negative (-7 + 2 = -5) So, this equation factored is (x + 2)(x - 7)
What is 2x squared times 3x squared?
2x2 x 3x2 = 6 x4 (2x)2 x (3x)2 = 36 x4
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.
Can you completely factor x32 - y32?
0
Can you please help me factor x4 plus 4y8?
It is: 1(x4+4y8) and can't be factored any further
You Need to factor X4-4?
It can be factored to: 4(x-1)
X4 plus 8x2 plus 4 - what are the factors?
(x2 + 4 - 2sqrt(3)) (x2 + 4 + 2sqrt(3))
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 x to the 4th power minus 49?
x4-49 is the difference of two squares when factored as (x2-7)(x2+7)
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
How do you factor the difference of two squares x8-5x2?
[ x8 - 5x2 ] is not the difference of two squares. [ 5x2 ] is not really a square.If you do have the difference of two squares, then:-- Take the square roots of the two terms.-- The factored form is (sum of the square roots) times (difference of the square roots)Example:Let's use . . . [ x8 - 9x2 ] .The square roots of the terms are . . . [ x4 ] and [ 3x ] .The factored form is (x4 + 3x) (x4- 3x).Go ahead and FOIL those factors out and you'll see.With this particular example, you could also continue, andfactor 'x' out of each of these factors, to wind up withx2 (x3 + 3) (x3 - 3)
How do you completely factor x to the fourth power minus ninety one?
You want to factor (x4 -91) First notice that the factors of 91 are 1, 7, 13, and 91. If we try them all , we see that x4 -91 is a prime polynomial. Even though the polynomial is prime, that is cannot be factored over the set of rational numbers, it is factorable over the set of irrational numbers. x4 - 91 = (x2)2 - (√91)2 = (x2 - √91)(x2 + √91) = [x2 - (√√91)2](x2 + √91) = (x - √√91)(x + √√91)(x2 + √91)
What is the simplest form of x to the 4th?
x^4 is in its simplest form.
X4 plus 7x2 minus 60 factor completely?
x4 + 7x2 - 60 = x4 + 12x2 - 5x2 - 60 = x2(x2 + 12) - 5(x2 +12) = (x2 - 5)(x2 +12)
How do you factor completely x to the fourth power minus 1?
So x4 - 1 is the difference of squares, so x4 - 1 = (x2 - 1)(x2 + 1) = (x + 1)(x - 1)(x2 + 1).
