What is the best way to factor x^4-16?
you can divid the probleme in two first x^4-16 you can write the 3rd form of principale relation (A^2-B^2 )=(A+B)(A-B) the difference of two squares above, which factors to (x^2 + 4)(x^2 - 4) You can't factor x^2 + 4 anymore, but x^2 - 4 is another difference of squares, exactly like the first factorise which can be factored to (x + 2)(x - 2), the final answer is (x^2 + 4)(x + 2)(x - 2)
How can I solve for [math]x[/math] when [math]x!/(x-3)! = 336[/math]?
Think of it in expanded form:[math]X!/(X-3)! [/math]is the same as:[math](1*2*3*...*(x-3)*(x-2)*(x-1)*x)/(1*2*3*...*(x-3))[/math]Now you can “cancel” everything from 1 through (x-3) from the top and the bottom. By “cancel”, I mean you can divide the top and bottom of this ratio by all the numbers from 1 through (x-3). Dividing the top and bottom by the same thing is essentially dividing by one, which doesn’t change the value of the expression. (I put “cancel” in quotes because I don’t like not acknowledging what you’re actually doing. When students skip over thinking about this, it trains them to think they can just cross off any pair of things that happen to look similar.) Now you have:[math]((x-2)*(x-1)*x)/1[/math]or[math](x-2)*(x-1)*x[/math]The three numbers being multiplied are some number which is 2 less than x, another number 1 less than x, and x. In other words, three consecutive whole numbers.At this point, since 336 isn’t very big, just pick some small x and see what happens. To save a bit of time, notice that 336 isn’t divisible by 5, so you can rule out any x that makes any of the three numbers be a multiple of 5.I’ll start with x=9[math]9*8*7=504[/math]Too large, but not too far off. Try x=8[math]8*7*6=336[/math]Bingo.
How do you execute the following polynomial division 5x^2-3x^2+6x-4 from x+2?
A Firing Squad?Processes to: Synthetic Subtraction? Any Takers?First thing you want to do is consolidate[math]5x^2-3x^2 to get 2x^2[/math]Observe the coefficient of xNext step is: Observe the coefficient of x and derive the factors. For 6, This is usually 2×3=6Observe the Polarity of the Last ArgumentNext step is: Figure out polarity going on in the last part of the trinomial. It is negative, this means:[math](2x-2)(x+2)=2x^2+(4x-2x)-4=2(x-1)(x+2)[/math]Without 4x=0 this may as well be a Pentagon.ExplanationOrdinarily you can utilize Synthetic Division on this.But in this scenario we really can't.Perfect Square or Diamond?Not to a Perfect Square You won't resolve, without a value for[math] 2x+x^2=-8[/math][math] [/math]via[math] [/math][math]x+2=-8/x[/math][math]PerfectSq=[x+2][x+2]=x^2+4x+4=64/x^2[/math]Perfect Rectangle? Or was it a Gong in a Chapeau?What Qualifies as a Rectangle? Certainly Not Mr. Alfred T. Bonkers the working class Rectangle. A Rectangle must be there.Please, Let me explain this Mr. Alfred T. Bonkers circumstance…In this there is Not Really a Rectangle that is there, but rather a Gone Rectangle, possibly just wed, got thirty seven empty coke cans tied to the rear bumper, big JUST MARRIED in soap on the rear window, nobody’s in the car, the car is parked at a gas station 3 miles from the hotel, he’s readying the do not disturb sign for the doorknob, she’s taking a bath, Tis a big night for The Newlyweds: The Bonkers, the Happy New Rectangle Couple, ‘Mr. And Mrs. Alfred T. Bonkers’ he Boasted as the Pair Checked-in, he in a wetsuit, ‘do you need a Phone out’ the Host inquired, ‘Maybe, if the Night doesn't go as planned, yeah, but I think I can unplug the phone myself thanks', said Alfred to the Admiration of his new wife, besides Alfred didn't want his friends, employer or family to know they Eloped nor really where he went, he left the office in a hurry, after seating a Gong with a Hat at his Desk at his Workplace in hopes nobody'd notice his 8 by 8 cubicle was totally vacated, we are the first persons to even notice this Gong in a Hat, we do wish we weren't, we may never live down the fact: Mr. Alfred T. Bonkers somehow slipped by we to make his exit.Any belief Otherwise, you won't resolve:[math]2x^2+6x-4=□[/math]Thus, without first landing a value for 6x+x^2=0, as better than x=-6…[math]Gong=[x-2][x+2]=x^2-4=32=(-8)(-4)[/math]Is most likely there.The hat is?[math]6x+x^2[/math]and seemingly somebody intends to swipe the hat.
How many real and imaginary roots (zeros of p(x)) are in the function p(x) =x^5+5x^3+17x^2+85? How?
There are 5 potential real zeroes and 5 imaginary roots. The number of imaginary roots will always be the degree of the polynomial. The maximum number of real roots will be the degree of the polynomial. How do we find this?The process through which we find the zeroes of the polynomial is as follows: First, use Descartes’ rule to find the number of positive and negative real zeroes. Then, use these to factor the polynomial, until it is all in the form of linear binomials and irreducible quadratics. Finally, factor the irreducible quadratics for imaginary zeroes or use the quadratic formula to find them. I will demonstrate below.First, we use Descartes’ Rule to find the number of positive and negative zeroes. We find the number of sign changes in the polynomial. If this number is even, the number of positive real zeroes is a positive even number less than or equal to this number. If it is odd, the number of positive real zeroes is a positive odd number less than or equal to this number. As we can see, there are no sign changes in the above polynomial(all of the coefficients are positive). To find the number of negative real zeroes, we apply the same rule, except instead of finding the number of sign changes in p(x), we find the number of sign changes in p(-x). This gives us: -x^5–5x^3+17x^2+85. Because there is one sign change(between -5x^3 and 17x^2), we conclude that there is one negative real zero. To find the real zeros of a polynomial, you must first find all of the factors of the constant(call that set p), and all of the factors of the leading coefficient(call that set q). The list of possible real rational zeroes will be all the possible combinations of p/q.Unfortunately, there is not a better way to find out which combination of p/q is the solution than to test them all until you find it; the only way is to plug them all in, or use synthetic division, etc. After you found the only real rational zero(c), factor the polynomial, such that it is in the form (x-c) times another polynomial(in this case, it will most likely be either quartic or cubic). Continue this process with the new polynomial, until the entire polynomial has been reduced to irreducible quadratics and linear binomials. Finally, find the complex zeroes by solving the irreducible quadratics either by factoring or by using the quadratic formula.I hope this answer helped!
Root is tan 30° and tan 15°of quadratic equation x^2+px+q, then what is 2+q-p?
Since, tan30° & tan15° are the roots of the given quadratic equation. Therefore, we have,-p = tan30° + tan15°& q = tan30°×tan15°We have to find, 2+q-p= 2+ tan30°×tan15° + tan30° + tan15°..…(1)Now, we know, tan45° = 1=> tan(30°+15°) = 1=> (tan30°+tan15°)/(1-tan30°×tan15°) =1=> tan30° + tan15° = 1- tan30°×tan15°=> tan30°×tan15° + tan30° + tan15° = 1Therefore, from (1) we get,2+q-p = 2+1 = 3.Hope, it'll help..!!P.S.- Feel free to comment if you have any query with it.Thank You!