Question

show that (x-2),(x+3) and (x-4) are factors of x3-3x power of 2-10x+24.

Answer 1

Answer:

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Answer 2

$\huge{\underline{\bigstar{\mathfrak{Solution!!}}}}$

$let \: p(x) = {x}^{3} - 3 {x}^{2} - 10x + 24$

$if\:(x-2) is \: a \: factor\: then \: p (2) = 0$

$p(2) = {2}^{3} - 3 ({2}^{2}) -10 (2) + 24$

$= 8 - 12 - 20 + 24$

$= 32 - 32 = 0$

$if\:(x+3) is \: a \: factor\: then \: p (-3) = 0$

$p (-3) = {-3}^{3} - 3 ({-3}^{2}) - 10 (-3) + 24$

$= - 27 - 27 + 30 + 24$

$= 54 - 54 = 0$

$if\:(x-4) is \: a \: factor\: then \: p (4) = 0$

$p (4) = {4}^{3} - 3 ({4}^{2} )- 10(4) + 24$

$= 64 - 48 - 40 + 24$

$= 84 - 84 = 0$

$\large{by \: factor \: theorem}}}}$

$\large\: { ( x-2), ( x+3) ,( x-4) are \: the \: factors \: of \: {x}^{3} - 3 {x}^{2} - 10x + 24$