Question

show that (x-2), (x-3) are the factors of x cube - 3x square - 10x + 24​

Answer 1

Answer:

p(x) = x³-3x²-10x+24

Given factors - (x-2) & (x-3)

x=2

x=3

If p(2) & p(3) = 0 ,they r factors of p(x)

p(2) = (2)³-3(2)²-10(2)+24 = 0

8-12-20+24 = 0

32-32 = 0

0 = 0

So, x-2 is a factor of p(x)

p(3) = (3)³-3(3)²-10(3)+24 = 0

27-27-30+24 = 0

51-57 = 0

-6 = 0

So, x-3 is not a factor of p(x)

Answer 2

$p(x) = {x}^{3} - {3x}^{2} - 10x + 24 \\ now \: \: g(x) = (x - 2 \: ) \\ g(x) = 0 \\ x - 2 = 0 \\ x = 2 \\ now \: put \: the \: value \: of \: {x = 2} \: \\ p(2) = {x}^{3} - {3x}^{2} - 10x + 24 \\ = {2}^{3} - {3 \times 2}^{2} - 10(2) + 24 \\ = 8 - 36 - 20 + 24 \\ = 32 - 56 = - 24 \\ \\ so \: (x - 2) \: is \: not \: a \: factor \: of \: p(x). \\ now \: g(x) = x - 3 \\ g(x) = 0 \\ x - 3 = 0 \\ x = 3 \\ \\ now \: \: put \: the \: value \: of \: {x \: = 3} \: \\ p(3) = {x}^{3} - {3x}^{2} - 10x + 24 \\ {3}^{3} - {3 \times 3}^{2} - 10(3) + 24 \\ = 27 - 18 - 30 + 24 \\ = 51 - 48 \\ = 3 \\ \\ so \: \: (x - 3) \: is \: not \: a \: factor \: of \: p(x). \\ \\ so \: (x - 2 )and \: (x - 3) \: are \: not \: the \: factors \: of \: p(x).$

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