Question

find the zeroes of the polynomial 2x^3-9x^2+x+12​

Answer 1

Step-by-step explanation:

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Given:

2x³ - 9x² + x + 12

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Solution:

Use hit and trial method..

take a factor x = -1

⇒ f(x)=-1

Substituting the equation we get..

⇒ f(1)= 2(-1)³ - 9(-1)² + (-1) + 12

⇒ -2+9-1+12

⇒ 9 - 9

⇒0

∴ (x+1) is a factor of 2x³ - 9x² + x +12

Now we divide 2x³ - 9x² + x +12 by (x+1)

         2x³ - 11x + 12

x+1) 2x³ - 9x² + x +12

      - 2x³ - 2x²

              -11x² +x

              +11x² +11x

                         12x + 12

                        -12x - 12

                                ×     

By dividing we get quadratic: 2x³ - 11x + 12

We find roots by solving for x;

x = [-b +(or) - √( b² - 4ac)]/ 2a

x = [ 11 +(or) - √ (121- 96)]/4

x = [ 11 +(or) - √25]/4

we get x =4 or x = 3/2;

therefor (x+1),(x-4) and (2x-3) are all factors of 2x³ - 9x² + x + 12.

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