factorise p(x)= x³+5x²-2x-24
Answers
Answer:
x³+5x²-2x-24
= x²(x+5) -2(x+12)
hope it helped :")
Answer:
Let’s let p(x) = x^3+5x^2–2x-24 represent your polynomial.
There is a relationship between values of x that make the polynomial 0 and factors of the polynomial (I’ll use that below).
If you’re allowed to use a graphing calculator then you can just graph the function and see where it crosses the x-axis. These *suggest* possible zeros of the polynomial (but you have to test them… to be certain.
Let’s assume that you’re not just going to graph it and look for potential zeros. You want test for the easy possibilities first, and to do that you’ll need to look at the constant term and the coefficient of the highest powered term. In this polynomial the constant term is (-24) the leading coefficient (1).
Any “easy” factors will come from fractions whose numerator (top) is a divisor of -24 and whose denominator is a divisor of (1). We don’t really care about the + or - bits— we’ll want to try them both anyway. That means we have +/- 1, +/- 2, +/- 3, +/- 4, +/- 6, +/- 12 and +/- 24 as possibilities:
The first thing I tried was plugging in 1:
p(1) = 1^3+6*1^2–2*1–24 which is NOT 0
Then I tried 2:
p(2)=2^3 +5*2^2 - 2*2–24 = 8+20–4–24=0
Since plugging in 2 makes the polynomial 0, I know that (x-2) is a factor (notice that if 2 makes the polynomial 0 then x-2 is the factor). I can use polynomial division (or something called “synthetic division” if you’re doing a lot of these sorts of problems) to figure out the next part:
This shows me that I can rewrite the polynomial as (x-2)(x^2+7x+12)
Now I want to work with the quadratic piece— You can use the same trick as we did above but it’s easy to just try factors of 12 and test them using FOIL (or use the quadratic formula to find the zeros directly). Of course, I could keep trying potential “easy” factors and I would find that
p(2) = 0
p(-4) = 0
p(-3) = 0
And from that I would also get the same result:
p(x) = (x-2)(x+4)(x+3)
How do you factor x^3 +2x^2 -5x -6?
What is the process to factorize p(x)=x3−2x2−5x+6 ?
How do I factor 2x3−7x2+2x+3 ?
How do I solve x^3+2x^2+5x+10=0?
How do you factor 2x^3+8x^2+5x+20?
x3+5x2−2x−24=x3+3x2+2x2−2x−24
=x2(x+3)+2x2−8x+6x−24
=x2(x+3)+2x(x−4)+6(x−4)=x2(x+3)+(x−4)(2x+6)
=x2(x+3)+2(x−4)(x+3)=(x+3)(x2+2(x−4))=(x+3)(x2+2x−8)
=(x+3)(x2+4x−2x−8)=(x+3)[x(x+4)−2(x+4)]
=(x+3)(x+4)(x−2)
Match with interesting people! Download Tinder now.
I would use a graphing calculator (ubiquitous in this level of math class these days) and look for “nice”, whole number x-intercepts. These x-values are also the zeros of the function and correspond directly to the factors of the polynomial.
Here is a snap shot from Desmos, since it is easier to embed here than a graphing calculator screen shot:
So, I see what appears to be a zero a -4, -3, and 2… these correspond to a polynomial function P(x) = a(x+4)(x+3)(x-2), where a will stretch, compress, or maybe even reflect the graph vertically, but it won’t change the zeros.
To solve for the a-value, we
x³+5x²-2x-24
First find the factors of 24:
24=2 x 2 x 2 x 3
Try:
(x+4)(x²+x-6)
Then:
(x+4)(x+3)(x-2)=x³+5x²-2x-24…………..
How do you factor x^3 +2x^2 -5x -6?
What is the process to factorize p(x)=x3−2x2−5x+6 ?
How do I factor 2x3−7x2+2x+3 ?
How do I solve x^3+2x^2+5x+10=0?
How do you factor 2x^3+8x^2+5x+20?
How do you factor 2x2+5x+3 ?
How can I factorize 3x2(x−1)2+2x(x−1)3 ?
How do you factor 2x^2-5x-3?
How do I factor 2x^2+7x+3?
How do I factorize x3+2x2+2x+1 ?
How do you factor 5x3−3x2+2x+6 ?
How do you factor 5x2(2x−3)13(3x+2)12+8x(2x−3)−23(3x+2)32
What is the factor of x3−3x2+2x ?
What is the value of x if x³ + x² + x + 1 =0?
How do I factor x^2 + x - 1?