Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

# What method do you use to factorise cubic equations?

For example, say you have x^3 + x^2 - 10x + 8, what are the steps you take to factorise it?

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• = x³ + x² - 10x + 8

= x³ + (2x² - x²) - (8x + 2x) + 8

= x³ + 2x² - x² - 8x - 2x + 8

= x³ + 2x² - 8x - x² - 2x + 8

= [x³ + 2x² - 8x] - [x² + 2x - 8]

= x.[x² + 2x - 8] - [x² + 2x - 8)

= (x - 1).[x² + 2x - 8]

= (x - 1).[x² + 2x + (1 - 1) - 8]

= (x - 1).[x² + 2x + 1 - 1 - 8]

= (x - 1).[(x² + 2x + 1) - (9)]

= (x - 1).[(x + 1)² - 3²] → you recognize: a² - b² = (a + b).(a - b)

= (x - 1).[(x + 1) + 3].[(x + 1) - 3)

= (x - 1).[x + 1 + 3].[x + 1 - 3)

= (x - 1).(x + 4).(x - 2)

• Usually trial and error with factors of 8 (+/-1, +/-2, +/-4)

if one works, then divide the main function by the corresponding factor:

like if x = 1 is a zero (root) then divide by (x-1)

This results in a quadratic, which is easier to factor (or use the quadratic formula)

• x^3 + x^2 - 10x + 8

= (x - 1) (x - 2) (x + 4)

• (1) By method of approximation to find the roots

first, get

f(x)=x^3+x^2-10x+8=(x+4)(x-2)(x-1)

(2) By the remainder theorem:

8 has the following factors:+/-(1,2,4,8) which

would be the rational roots if the equation has

any. Try x=1 get

1 +1 -10 +8 | 1

...+1 + 2 - 8 |

------------------

1 + 2 - 8

=>

f(1)=0=>

f(x)=(x-1)(x^2+2x-8)=>

f(x)=(x-1)(x-2)(x+4)

• See if you like this method, (where integer results expected).

y = x^3 + x^2 - 10x + 8

Factors of the constant 8 are plus or minus 1, 2, 3

If substituting a constant makes y zero you have found a factor.

Trying the least value, x = 1, first.

1 + 1 – 10 + 8 = 0, a result. Then write in a half solved quadratic

y = (x – 1)(x^2 + kx – 8) = x^3 + x^2 - 10x + 8

coefficients of x: -k - 8 = -10, so, k was 2,

x^2 + 2x – 8 = (x – 2)(x + 4)

• My first step is to try the rational root theorem.  There's no guarantee that it'll work, but it's worth a try.  There are methods, like Vieta's Formula, Cardona's Formula, and using the method for depressed cubic for Cubics, but it plays hell trying to use them.

Rational root theorem

x^3 + x^2 - 10x + 8 = 0

1 is divisible by -1 and 1

8 is divisible by -8 , -4 , -2 , -1 , 1 , 2 , 4 , 8

Since the leading coefficient is 1, the divisors will just be the divisors of 8

x = 1 looks like it will work

1^3 + 1^2 - 10 * 1 + 8 = 1 + 1 + 8 - 10 = 10 - 10 = 0

x = 1 works so x - 1 = 0 is a solution, which means that (x - 1) is a factor

At this point, you can try another root or you can find the quadratic that gives us the cubic when multiplied by x - 1

(x - 1) * (ax^2 + bx + c) = x^3 + x^2 - 10x + 8

We know that a = 1

(x - 1) * (x^2 + bx + c) = x^3 + x^2 - 10x + 8

x^3 + bx^2 + cx - x^2 - bx - c = x^3 + x^2 - 10x + 8

x^3 = x^3

bx^2 - x^2 = x^2

cx - bx = -10x

-c = 8

-c = 8, therefore c = -8

c - b = -10

-8 - b = -10

-8 + 10 = b

2 = b

x^2 + 2x - 8 is the quadratic we're looking for

x^2 + 2x - 8 = 0

x^2 + 2x = 8

x^2 + 2x + 1 = 8 + 1

(x + 1)^2 = 9

x + 1 = -3 , 3

x = -3 - 1 , 3 - 1

x = -4 , 2

x = -4 , 1 , 2 are your roots.

• x³ + x² - 10x + 8

If there is a rational root, it would be a factor of the constant term over a factor of the high-degree coefficient.  Since the latter is 1, we only have to look at the constant.

Here is the list of possible rational roots based on the above information:

±1, ±2, ±4, ±8

There are 8 possible outcomes.  Now we brute-force this list until we find one.  If we find one we can turn it into a factor and divide the original polynomial by that factor to get the remaining quadratic.  At that point we have many ways to find the roots, even if they aren't rational.

Testing the 1s looking for zeroes:

x³ + x² - 10x + 8

(-1)³ + (-1)² - 10(-1) + 8 and 1³ + 1² - 10(1) + 8

-1 + 1 + 10 + 8 and 1 + 1 - 10 + 8

18 and 0

We have a zero at x = 1 so (x - 1) is a factor.  Now we can divide to get the remaining quadratic:

. . . . _x²_+_2x_-_8___

x - 1 ) x³ + x² - 10x + 8

. . . . . x³ - x²

. . . . -------------

. . . . . . . . 2x² - 10x + 8

. . . . . . . . 2x² - 2x

. . . . . . . . ------------

. . . . . . . . . . . . -8x + 8

. . . . . . . . . . . . -8x + 8

. . . . . . . . . . . . ------------

. . . . . . . . . . . . . . . . . 0

Now we have:

(x - 1)(x² + 2x - 8)

I can see this can be factored further:

(x - 1)(x + 4)(x - 2)

Resulting in the roots being:

x = -4, 1, 2

And notice that all three roots are in my list of 8 possible rational roots.

• Usually, I just put it into a computer program. Or I might graph it, then use Newton's method. In this case, a graph shows roots near x = -4, 1, and 2. Trying them out, it seems those values work perfectly. So the factors are (x+4), (x -1), and (x -2).