Question

show that (x-2),(x+3),(x-4)are factor of x³-3x²-10x+24​

Answer 1

Answer:

X+a=x-2

p(2)=(2)^3-3(2)^2-10(2)+24

=8-12-20+24

=24-24

=0

p(-3)=(-3)^3-3(-3)^2-10(-3)+24

=-27-27+30+24

=-54+54

=0

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

=64-48-40+24

=88-88

=0

therefore these are all factors of x^3-3x^2-10x+24

Answer 2

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

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

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

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

$= 8 - 12 - 20 + 24$

$= 32 - 32 = 0$

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

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

$= - 27 - 27 + 30 + 24$

$= 54 - 54 = 0$

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

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

$= 64 - 48 - 40 + 24$

$= 84 - 84 = 0$

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

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