Factorise 2x^3 - 9x^2 + x + 12
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The answer to your question is typed below↓... ____________________________________________ Given: 2x³ - 9x² + x + 12 ____________________________________________ 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. ____________________________________________
Answer: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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