Help factorise cubic?!?
Use the formula A³+B³ = (A+B)(A²-AB+B²) so here we have A=x and B = 3 (27=3³) => x³+27 = x³ + 3³ = (x+3)(x²-3x+9)
Help with cubic function and distinct zeros?
If the polynomial has a zero at x=2 then (x-2) is a factor. Likewise (x-5) is a factor. That only leaves one root to determine, call that root r. Then the polynomial must have the form p(x) = a(x-2)(x-5)(x-r) where a is a constant. To get a maximum at x=5, we expand and calculate p'(x) p'(x) = a [ 3x² -2(r+7) x +7r + 10] p'(5) = 3a (5-r) which has to be zero for a minimum at x=5. This means r=5. Now to find out what will make this a local maximum. We analyse the second derivative. With r=5 p'(x) = a (3x^2 - 24x + 45) = 3a (x^2-8x+15) p"(x) = 3a (2x-8) So p"(5) = 6a For x=5 to be a maximum, p"(5) < 0, which will be the case if a is any negative number. Choose a=-1. Solution: p(x) = -(x-2)(x-5)^2 [ConMan's argument is a good one, too.]
What is the process of factoring a cubic function?
I will show you two fool-proof methods to factorise a cubic.
How do you factor out cubic functions?
Firstly you have to realise that not all cubic functions will factorise easily into factors with real numbers. I don’t think you are talking about “complex number” factors!This is similar to what we do with Quadratics.So, if you are talking about cubics which have REAL factors thenI will show you two fool-proof methods to factorise them.This one is a great example:You need to start by finding a factor as follows...Just try substituting some simple numbers like 1 or 2 or 3 or perhaps – 1 or – 2 etc until f(x) comes to zero.In this case try x = 1 then f(1) = 1 – 2 – 1 + 2 = 0————————————————————————————————————METHOD 2 (This involves good logical thinking. I like really it!)Once you get a good grasp of the logic, all you need to write is Step 4!You do the rest mentally.========================================================
How do you factor cubic equations?
I assume you are asking to factorize the left hand side of an equation of the form[math]ax^3+bx^2+cx+d=0.[/math]The answer is that it depends on whether you want to reduce your cubic polynomial over [math]\mathbb{Q}[/math], over [math]\mathbb{R}[/math] or over [math]\mathbb{C}[/math].If you want to factorize over [math]\mathbb{Q}[/math], you could make use of the Factor theorem, coupled with the fact that the only rational solutions of the equation above are of the form [math]\frac{p}{q}[/math], where [math]p[/math] is a factor of [math]d[/math] and [math]q[/math] is a factor of [math]a[/math]. If none of these rational numbers satisfy the above equation, then the cubic expression on the L.H.S. is irreducible over [math]\mathbb{Q}[/math]; otherwise, factor out the linear factor [math](qx - p)[/math], use polynomial division, and try to factorize the resulting quadratic using this same technique.If you want to factorize over [math]\mathbb{C}[/math], you could use the Cubic Formula illustrated in the link. You can use the same formula to factorise over [math]\mathbb{R}[/math], except that if you get any complex roots, you'll need to multiply out their linear factors again.Detailed example:[math]2x^3+9x^2-13x+4[/math]Possible rational solutions of the equation [math]2x^3+9x^2-13x+4=0[/math] are [math]\pm 1, \pm 2, \pm 4[/math] and [math]\pm \frac{1}{2}[/math]. Since [math]\frac{1}{2}[/math] is a solution [math](2(\frac{1}{2})^3+9(\frac{1}{2})^2-13(\frac{1}{2})+4=0)[/math], we factor out [math](2x-1)[/math] and divide, getting [math]2x^3+9x^2-13x+4 = (2x-1)(x^2+5x-4)[/math]. We repeat the process for the resulting quadratic [math]x^2-5x+4[/math], which, however, is irreducible over [math]\mathbb{Q}[/math]. So, over [math]\mathbb{Q}[/math], the cubic function [math]2x^3+9x^2-13x+4[/math] factorizes to [math](2x-1)(x^2+5x-4)[/math].Over [math]\mathbb{R}[/math], however, the quadratic [math]x^2+5x-4[/math] factorizes into [math]\left(x+\frac{5-\sqrt{41}}{2}\right)\left(x+\frac{5+\sqrt{41}}{2}\right)[/math], so [math]2x^3+9x^2-13x+4[/math] factorizes to [math](2x-1)\left(x+\frac{5-\sqrt{41}}{2}\right)\left(x+\frac{5+\sqrt{41}}{2}\right)[/math] over [math]\mathbb{R}[/math] (and over [math]\mathbb{C}[/math]).Quick example:[math]x^3+3x[/math]This cubic factorizes into [math]x(x^2+3)[/math] over [math]\mathbb{Q}[/math] and over [math]\mathbb{R}[/math], but factorizes into [math]x(x+i\sqrt{3})(x-i\sqrt{3})[/math] over [math]\mathbb{C}[/math].
What are the x and y intercepts for a cubic function?
Set x=0 to find the y intercept and set y=0 to find x intercept. I'll demonstrate with an example. Let's assume: y(x) = x^3 - 2x^2 - 5x + 6 To get the y-intercept: Set x=0: y(0) = (0)^3 - 2(0)^2 - 5(0) + 6 Therefore y(0)=6 To get the x-intercepts: Set y=0: 0 = x^3 - 2x^2 - 5x + 6 Now we have to factorise, guess a couple of values for x, for instance x=1 will result in y=0. So this leaves us with 0 = (x-1)(x^2 - x - 6) Factorising the second bracket into (x + 2)(x - 3) Therefore 0 = (x-1)(x+2)(x-3) So the x intercepts (also known as the roots of the equation) are: X=1 X=-2 and X=3 I hope this explanation will help you with this and future problems :)
Factorising a cubic polynomial?
Question Number 1 For this polynomial equation x^3 -x^2 -65*x -63 = 0, answer the following questions : A. Solve by Factorization Answer For Question 1 x^3 -x^2 -65*x -63 = 0 And we get P(x)=x^3 -x^2 -65*x -63 Now, we can look for the roots of P(x) using various Algorithm : 1A. Solve by Factorization x^3 -x^2 -65*x -63 = 0 Separate : ( x^3 -9*x^2 ) + ( 8*x^2 -72*x ) + ( 7*x -63 ) = 0 Commutative Law : ( x^3 +8*x^2 +7*x ) + ( -9*x^2 -72*x -63 ) = 0 Distributive Law : x*( x^2 +8*x +7 ) + -9*( x^2 +8*x +7 ) = 0 Factor : ( x -9 )*( x^2 +8*x +7 ) = 0 Separate : ( x -9 )*( ( x^2 +x ) + ( 7*x +7 ) ) = 0 Commutative Law : ( x -9 )*( ( x^2 +7*x ) + ( x +7 ) ) = 0 Distributive Law : ( x -9 )*( x*( x +7 ) + +*( x +7 ) ) = 0 Factor : ( x -9 )*( x +1 )*( x +7 ) So the Polynomial have 3 roots : x1 = 9 x2 = -1 x3 = -7
How to factor quintic functions?
This is on our final review and it was never even taught in the course. fml. The function is: f(x) = x^5 - 2x^4 + 8x^2 - 13x + 6 I get how to find the zeroes of this which are 1, 1, and -2. So that at least I know it factors part of it to (x-1)^2 * (x+2) and I *think* that the last part would be a trinomial whose quadratic function would yield the remaining roots... online calcs show me this is x^2 - 2x + 3 but I have no clue how to get there. Could anyone show me the step by step on how to factor that equation into (x-1)^2 * (x+2) * (x^2 - 2x + 3)? Please? I really need to understand this and nowhere in the textbook do they tell you how to do anything more than cubic functions. :'(
Factorising Cubic Polynomial?
.If you hate synthetic division (long division ) try the following: (x - 3)(ax² + bx + c ) = 4x³ - 12x² + 2x - 6 ax³ + (b - 3a)x² + (c - 3b)x - 3c = 4x³ - 12x² + 2x - 6 Like-terms together a = 4 b - 3a = -12 ━━━━➤ b = 3a - 12 = 3(4) - 12 = 0 ∴ b = 0 c - 3b = 2 ∴ c = 2 ax² + bx + c = 4x² + 2 = 2(2x² + 1) Combine: 4x³ - 12x² + 2x - 6 = 2(x - 3)(2x² + 1) ━━━━━━━━