Question

2 Using factor theorem, factorize x3-2x2-5x+6

Answer 1

• (x-1) ( x+2) (x+3) is the answer
• explanation
• (x=1 )
• 1^3-2(1)^2 - 5 (1) + 6
• 1-2-5+6
• -1+1
• R= 0
• further solve it
• (divide )

Answer 2

HEY MATE HERE IS YOUR ANSWER

${x}^{3} - 2 {x}^{2} - 5x + 6$

factors of 6 are = 1,2,3,6,-1,-2,-3,-6

lets put x =1 in p(x)

$p(1) = {1}^{3} - 2 ({1})^{2} - 5(1) + 6 \\ \: \: \: \: \: \: \: \: \: = 1 - 2 - 5 + 6 \\ \: \: \: \: \: \: \: \: \: = 7 - 7 \\ \: \: \: \: \: \: \: \: \: = 0$

so, x=1

x-1 =0

(x-1) is factor

now divide p(x) by (x-1)

$we \: get \: {x}^{2} - x - 6$

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

so factors are (x-1), (x-3), (x+2)

I HOPE IT HELPS AND IF IT DOES PLEASE NARK AS BRAINLIEST ANSWER

AND THANKS AND ALSO RATE 5 STAR

STAY HOME STAY SAFE.