Doo 2 3 3 X 4

We deep dive into why the Sea-Doo SPARK is in a class of its own and has Re-Sparked the watercraft industry with creative design concepts and masterful engin. mathx^4–4x^3+4x^2+4(x^2–2x+1)=0/math math(x(x-2))^2 +4(x-1)^2=0/math math(x^2–2x)^2+ 4(x-1)^2=0/math math(x-1)^2–1 ^2+4(x-1)^2=0/math math.

• Doo 2 3 3 X 4 4
• Doo 2 3 3 X 4 X 6 4
• 4 3 X 1
• X 1 X2 X 6 2 4

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Usually, simple polynomial factoring will be, well, fairly simple. However, there are instances when the factoring will, in a technical sense, be 'simple' (because all you're doing is taking a factor, common to all of the terms, out front), the factoring will, in an actual sense, be messy (because that common factor will be complex or large, or because there are loads of terms to consider).

The only difference, really, will be in the care one needs to take — along with perhaps needing an application of the formal process for finding the GCF.

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• Factor 3x³ + 6x² – 15x.

I can factor a 3 and an x out of each term:

3x³ = 3x(x²)

6x² = 3x(2x)

–15x = 3x(–5)

Being careful of my signs, I factor:

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• Factor 26x sqrt[9y³] – 13y sqrt[4x²y] + xy sqrt[169x⁴y³], assuming that all variables are non-negative.

At first glance, it doesn't look like anything is common to all three terms. But then I remember that I should probably see if I can simplify the square roots first. I'll work with each of the terms separately.

First term:

sqrt[9y³]

= sqrt[9] sqrt[y²] sqrt[y]

= 3 × y × sqrt[y]

26x sqrt[9y³]

= (26x) × 3y sqrt[y]

= 78xy sqrt[y]

Second term:

sqrt[4x²y]

= sqrt[4] sqrt[x²] sqrt[y]

= 2 × x × sqrt[y]

13y sqrt[4x²y]

= (13y) × 2x sqrt{y]

= 26xy sqrt[y]

Third term:

sqrt[169x⁴y³]

= sqrt[169] sqrt[x⁴] sqrt[y³]

= 13 × x² × y sqrt[y]

xy × sqrt[169x⁴y³]

= 13x³y sqrt[y]

Comparing the three terms, I see... well, I see that I have a mess. Clearly, there's a common factor of sqrt[y] in each of the three terms, but I think I'll turn to the formal method for finding the GCF of the rest of each of the three terms:

Okay, so my common factor, in addition to the square root, will be 13xy. What is left from each of the terms (to go inside the parentheses)? I'll list these out nicely, too, working from what I did above:

With this, I can do my factorization. As I go, I'll be careful to insert the correct signs between the terms.

13xy sqrt[y](6 – 2 + x²)

Huh. I hadn't even noticed that I could combine two of the terms. But the 6 and the 2 will obviously simplify. My final answer then is:

13xy sqrt[y](4 + x²)

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• Factor2(x – y) – b(x – y).

This expression may seem completely different from what I've done before, but really it's not. The two terms, 2(x – y) and –b(x – y), do indeed have a common factor; namely, the parenthetical factor x – y. This binomial may be different from what I'm used to seeing referred to as being a 'factor', but the factorization process works just the same for this expression as it did for every other expression before.

First, I'll take the common factor out front, and then I'll draw an open-paren:

2(x – y) – b(x – y)

From the first term, I have a 2 left over; this starts my parenthetical:

2(x – y) – b(x – y)

From the second term, I have a '–b' left over; this finishes my parenthetical:

2(x – y) – b(x – y)

I've ended up with the factored form being the product of two binomials. This is the 'undo' of what is commonly referred to as 'FOILing', or multiplying two binomials. My final answer is:

Doo 2 3 3 X 4 4

(x – y)(2 – b)

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• Factor x(x – 2) + 3(2 – x).

This is almost the same as the previous case, but not quite, because the 'x – 2' factor in the first term is not quite the same as the '2 – x' in the second term. But they're almost the same.

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If the parentheticals had been 'x + 2' and '2 + x', the factors would have been the same; that is, I could have reversed the terms in one of the factors, to make it match the other factor, because order doesn't matter for addition.

But order does matter for subtraction, so I don't actually have a common factor here. But I would have a common factor if I could just flip (that is, reverse the order of) that subtraction.

What happens when we reverse a subtraction? To figure this out, take a look at the following numerical subtractions:

The subtraction in the second line is reversed from the subtraction in the first line. As a result, we got the same answer except that the sign had changed. We reversed the subtraction, and got the previous answer, but with the opposite sign.

This is always true: When you flip a subtraction, you also flip the sign out front.

(Note: When reversing a subtraction, it can be helpful to put parentheses around the new subtraction, with a 'minus' sign out front.)

For the expression they've given me, flipping the subtraction results in:

x(x – 2) + 3(2 – x)

By reversing the subtraction in the second parenthetical, I have created a common factor. Now I can proceed as I had in the previous example:

x(x – 2) + 3(2 – x)

These last two examples lead us to the next topic: factoring 'in pairs'.

URL: https://www.purplemath.com/modules/simpfact2.htm

How do you factor # x^4 + x^3 + x^2 + x + 1#?

1 Answer

#x^4+x^3+x^2+x+1#

#=(x^2+(1/2+sqrt(5)/2)x+1)(x^2+(1/2-sqrt(5)/2)x+1)#

Explanation:

This quartic has four zeros, which are the non-Real Complex #5#th roots of #1#, as we can see from:

#(x-1)(x^4+x^3+x^2+x+1) = x^5-1#

So if we wanted to factor this polynomial as a product of linear factors with Complex coefficients then we could write:

#x^4+x^3+x^2+x+1#

#=(x-(cos((2pi)/5) + i sin((2pi)/5))) * (x-(cos((4pi)/5) + i sin((4pi)/5))) * (x-(cos((6pi)/5) + i sin((6pi)/5))) * (x-(cos((8pi)/5) + i sin((8pi)/5)))#

A cleaner algebraic approach is to notice that due to the symmetry of the coefficients, if #x=r# is a zero of #x^4+x^3+x^2+x+1#, then #x=1/r# is also a zero.

Hence there is a factorisation in the form:

#x^4+x^3+x^2+x+1#

#=(x-r_1)(x-1/r_1)(x-r_2)(x-1/r_2)#

#=(x^2-(r_1+1/r_1)x+1)(x^2-(r_2+1/r_2)x+1)#

So let's look for a factorisation:

#x^4+x^3+x^2+x+1#

#=(x^2+ax+1)(x^2+bx+1)#

#=x^4+(a+b)x^3+(2+ab)x^2+(a+b)x+1#

Equating coefficients we find:

#a+b = 1#

#2+ab=1#, so #ab = -1# and #b=-1/a#

Substituting #b=-1/a# in #a+b=1# we get:

Doo 2 3 3 X 4 X 6 4

#a-1/a = 1#

Hence:

#a^2-a-1 = 0#

Using the quadratic formula, we can deduce:

#a = 1/2 +- sqrt(5)/2#

Since our derivation was symmetric in #a# and #b#, one of these roots can be used for #a# and the other for #b#, to find:

#x^4+x^3+x^2+x+1#

#=(x^2+(1/2+sqrt(5)/2)x+1)(x^2+(1/2-sqrt(5)/2)x+1)#

4 3 X 1

If we want to factor further, use the quadratic formula on each of these quadratic factors to find the linear factors with Complex coefficients.

X 1 X2 X 6 2 4

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