Question

factorise 2x^3-11x^2+17x-6 using factor theorem​

Answer 1

$Let \: p(x) = 2x^{3} - 11x^{2} + 17x - 6$

$By \: trial , we \: find \: that \: p(2) = 0$

$i.e., p(2) = 2\times 2^{3} - 11\times 2^{2} + 17\times 2 - 6$

$= 2 \times 8 - 11 \times 4 + 34 - 6$

$= 16 - 44 + 34 - 6$

$= 50 - 50$

$= 0$

$\therefore (x-2) \: is \: a \: factor \: of \: p(x)$

x-2| 2x³-11x²+17x-6|2x²-7x+3

***** 2x³-4x²

_______________

********** -7x²+17x

********** -7x² +14x

________________

************** 3x - 6

************** 3x - 6

________________

*************** (0)

$p(x) = (x-2)(2x^{2}-7x+3)$

$= (x-2)(2x^{2} - 6x - 1x + 3 )$

$=(x-2)[ 2x(x-3)-1(x-3)]$

$= (x-2)(x-3)(2x-1)$

Therefore.,

$\red{ Factors \: of \: 2x^{3} - 11x^{2} + 17x - 6}$

$\green { = (x-2)(x-3)(2x-1)}$

•••♪

Answer 2

ANSWER= (x-2)(x-3)(2x-1)

For rxplanation pls view the shared jpeg