Home › Forums › Math › One root of f(x) = 2x^3 + 9x^2 + 7x – 6 is –3. explain how to find the factors of the polynomial. › Reply To: One root of f(x) = 2x^3 + 9x^2 + 7x – 6 is –3. explain how to find the factors of the polynomial.

One root of f(x) = 2x^{3} + 9x^{2} + 7x – 6 is –3 this means

(x+3) is a factor of polynomial.

So, to find we will try to reduce this cubic into a quadratic equation by using this given factor.

To do that we have to take out x+3 common out from the equation

f(x) = 2x^{3} + 9x^{2} + 7x – 6

Splitting the x^{2} and x term and getting x+3 common

=> 2x^{3} + 6 x^{2} + 3 x^{2} + 9x – 2x – 6

=> 2x^{2} .( x + 3) + 3x .(x + 3) -2 .(x +3)

=> ( x+3) .( 2x^2 + 3x -2 )

Now factorizing the quadratic equation as usual.

=> (x +3) ( 2 x^{2} + 4x – x – 2)

=> ( x +3) [ 2x (x + 2) – ( x+ 2)]

=> ( x + 3 )( x + 2) ( 2x – 1 ) = 0

x + 2 = 0 or 2x – 1 = 0

So, the other factors are -2 and 1/2

**Note: **If you are not able to take the given factor common as shown above, you can also use polynomial division to get the quadratic.

- This reply was modified 2 years ago by ProtonsTalk.
- This reply was modified 2 years ago by ProtonsTalk.