Question

Find factors of x³+12x²-19x= -20

Answer 1

`\boxed{\mathrm{No \: factors}}`

Explanation:

`x^3 +12x^2 -19x= -20`

Add 20 on both sides.

`x^3 +12x^2 -19x+20= 0`

Add 67x and 44 on both sides.

`x^3 +12x^2 +48x+64= 67x + 44`

Factor left side of the equation.

`(x+4)(x+4)(x+4)= 67x + 44`

`(x+4)^3=67x+44`

Subtract 67x and 44 on both sides.

`(x+4)^3-67x-44=0`

Cannot be factored further.

This is a prime expression. A prime expression cannot be factored.

We cannot factor out x from the expression, there are no factors.

Answer 2

No Factors

Explanation:

`x^3-12x^2-19x+20 = 0\\Let \ p(x) = x^3-12x^2-19x+20`

Factors of 20 = ±1, ±2 , ±4 , ±5 , ±10 and ±20

±1, ±2 and ±3 are not the factors of given polynomial.

Putting x = 4 in the given polynomial

`p(4) = (4)^3+12(4)^2-19(4)+20\\p(4) = 64+192-76+20\\p(4) = 200`

So, x = 4 is not a factor of p(x)

Putting x = -4 in the given equation

`p(-4) = (-4)^3+12(-4)^2-19(4)+20\\p(-4) = -64+192-76+20\\p(-4) = 73`

So, x = -4 in the given equation

Putting x = 5 in the given equation

`p(5) = (5)^3+12(5)^2-19(5)+20\\p(5) = 125+300-95+20\\p(5) = 350`

So, x = 5 is not a factor of p(x)

Putting x = -5 in the given equation

`p(-5) = (-5)^3+12(-5)^2-19(-5)+20\\p(-5) = -125+300+95+20\\p(-5) = 290`

So, x = -5 is not a factor of p(x)

Putting x = 10 in the given equation

`p(10) = (10)^3+12(10)^2-19(10)+20\\p(10) = 1000+1200-190+20\\p(10) = 2030`

So, x = 10 is not a factor of p(x)

Putting x = -10 in the given equation

`p(-10) = (-10)^3+12(-10)^2-19(-10)+20\\p(-10) = -1000+1200+190+20\\p(-10) = 410`

So, x = -10 is not a factor of p(x)

Putting x = 20 in the given equation

`p(20) = (20)^3+12(20)^2-19(20)+20\\p(20) = 8000+4800-380+20\\p(20) = 12440`

So, x = 20 is not a factor of p(x)

Putting x = -20 in the given equation

`p(-20) = (-20)^3+12(-20)^2-19(-20)+20\\p(-20) = -8000+4800+380+20\\p(-20) = -2800`

So, x = -20 is not a factor of p(x)

From the above solution, we conclude that the given equation can not be factorized and thus, has no factors.