Question

factorise 2x3-9x2+x+12

Answer 1

Greetings,
The answer to your question is typed below↓...
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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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Ps.: Enjoy ;)
       

Answer 2

Answer:

Answer is given below with a pic

Step-by-step explanation:

The factors of the cubic equation 2x³-9x²+x+12 are (x-4)(2x-3)(x+1).

We can simply find the one root of the cubic equation 2x³-9x²+x+12 by using the hit and trial method.

                           p(x)= 2x³-9x²+x+12

                         p(-1) =2(-1)³-9(-1)²+(-1) +12

                                = -12 +12

                                = 0

So, (x+1) is one factor of 2x³-9x²+x+12.

By using synthetic division, other factors can be determined,

Depressed equation = 2x²-11x+12

                                 = 2x² - 8x-3x+12

                                  = 2x(x-4)-3(x-4)

                                 = (x-4)(2x-3)

Hence, the factors of 2x³-9x²+x+12 are (x-4)(2x-3)(x+1).