Question

factorise 2y3 - 11y2+17y-6​

Answer 1

$\huge\boxed{\underline{\mathcal{\red{A}\green{N}\pink{S}\orange{W}\blue{E}\pink{R}}}}$

${\bf{\red{=>2y^3-11y^2+17y-6}}}$

${\bf{\blue{=>(y-3)(2y^2-5y+2)}}}$

${\bf{\pink{=>(y-3)(2y^2-4y-y+2}}}$

${\bf{\green{=>(y-3)(2y(y-2)-1(y-2)}}}$

${\bf{\orange{=>(y-3)(2y-1)(y-2)}}}$

Answer 2

=>2y 3 −11y 2 +17y−6

(y-3)(2y^2-5y+2(y−3)(2y 2−5y+2)

(y-3)(2y^2-4y-y+2) (y−3)(2y 2−4y−y+2)

(y-3)(2y(y-2)-1(y-2)(y−3)(2y(y−2)−1(y−2)

(y-3)(2y-1)(y-2)(y−3)(2y−1)(y−2)