How to find factors or roots of a cubic polynomial!
Urgent help required!
mysticd:
Polinomials doesn't have roots .we can find zeroes of the cubic polynomials using Harnors method
Or factor theorem
I need zeroes of the polynomial
do have any problem , i can explain it
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Group the polynomial into two sections. Grouping the polynomial into two sections will let you attack each section individually.Say, "We're working with the polynomial." x3 + 3x2 - 6x - 18 = 0. Let's group it into (x3 + 3x2) and (- 6x - 18)2Find what's the common in each section.Looking at (x3 + 3x2), we can see that x2 is common.Looking at (- 6x - 18), we can see that -6 is common.3Factor the commonalities out of the two terms.Factoring out x2 from the first section, we get x2(x + 3).Factoring out -6 from the second section, you'll get -6(x + 3).4If each of the two terms contains the same factor, you can combine the factors together.This gives you (x + 3)(x2 - 6).5Find the solution by looking at the roots. If you have an x2 in your roots, remember that both negative and positive numbers fulfill that equation.The solutions are -3, √6 and -√6.Part2Factoring Using the Free Term1Rearrange the expression so it's in the form of aX3+bX2+cX+d.Let's say you're working with the equation: x3 - 4x2 - 7x + 10 = 0.2Find the all of the factors of "d". The constant "d" is going to be the number that doesn't have any variables, such as "x," next to it.Factors are the numbers you can multiply together to get another number. In your case, the factors of 10, or "d," are: 1, 2, 5, and 10.3Find one factor that causes the polynomial to equal to zero. We want to determine which factor makes the polynomial equal zero when we substitute the factor for each "x" in the equation.Start by using your first factor, 1. Substitute "1" for each "x" in the equation:
(1)3 - 4(1)2 - 7(1) + 10 = 0This gives you: 1 - 4 - 7 + 10 = 0.Because 0 = 0 is a true statement, you know that x = 1 is a solution.4Do a little rearranging. If x = 1, you can rearrange the statement to look a bit different without changing what it means."x = 1" is the same thing as "x - 1 = 0" or "(x - 1)". You've just subtracted a "1" from each side of the equation.5Factor your root out of the rest of the equation. "(x - 1)" is our root. See if you can factor it out of the rest of the equation. Take it one polynomial at a time.Can you factor (x - 1) out of the x3? No you can't. But you can borrow a -x2 from the second variable; then factor it: x2(x - 1) = x3 - x2.Can you factor (x - 1) out of what remains from your second variable? No, again you can't. You need to borrow another little bit from the third variable. You need to borrow a 3x from -7x. This gives you -3x(x - 1) = -3x2 + 3x.Since you took a 3x from -7x, our third variable is now -10x and our constant is 10. Can you factor this? You can! -10(x - 1) = -10x + 10.What you did was rearrange the variables so that you could factor out a (x - 1) out of the entire equation. Your rearranged equation looks like this: x3 - x2 - 3x2 + 3x - 10x + 10 = 0, but it's still the same thing as x3 - 4x2 - 7x + 10 = 0.6Continue to substitute by the factors of the free term. Look at the numbers that you factored out using the (x - 1) in Step 5:x2(x - 1) - 3x(x - 1) - 10(x - 1) = 0. You can rearrange this to be a lot easier to factor one more time: (x - 1)(x2 - 3x - 10) = 0.You're only trying to factor (x2 - 3x - 10) here. This factors down into (x + 2)(x - 5).7Your solutions will be the factored roots. You can check whether your solutions actually work by plugging each one, individually, back into the original equation.(x - 1)(x + 2)(x - 5) = 0 This gives you solutions of 1, -2, and 5.Plug -2 back into the equation: (-2)3 - 4(-2)2 - 7(-2) + 10 = -8 - 16 + 14 + 10 = 0.Plug 5 back into the equation: (5)3 - 4(5)2 - 7(5) + 10 = 125 - 100 - 35 + 10 = 0.
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Heya dear
let a function in cubic polynomial in x
•First of all use hit and trial method and put ± values of x respectively generally the values satisfying the equation does not exeeds 3 or -3
• then let the 1st value of x is 1 so (x-1) will be the factor
• now divide the cubic polynomial with (x-1)
you will get a quotient in quadratic equation.
• now factorise the quadratic polynomial .
• you will get all the there factors or if you put them equal to 0 then you will get the zeros(roots)
do ask me if you have any problem .
let a function in cubic polynomial in x
•First of all use hit and trial method and put ± values of x respectively generally the values satisfying the equation does not exeeds 3 or -3
• then let the 1st value of x is 1 so (x-1) will be the factor
• now divide the cubic polynomial with (x-1)
you will get a quotient in quadratic equation.
• now factorise the quadratic polynomial .
• you will get all the there factors or if you put them equal to 0 then you will get the zeros(roots)
do ask me if you have any problem .
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