Question

3. Show that (x - 2), (x + 3) and (x - 4) are factors of x-3x - 10x + 24.​

Answer 1

Step-by-step explanation:

x-2=0

x=2

so

p(x)=x-3x-10x+24

p(2)=2-6-20+24

=26-26

=0. therefor,x-2 is a factor

x+3 =0

x=-3

p(-3)= -3+9+30+24

=60. therefore,x+3 is not a factor

x-4=0

x=4

p(4)=4-12-40+24

=-24. therefor,x-4 is not a factor

Answer 2

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

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

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

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

$= 8 - 12 - 20 + 24$

$= 32 - 32 = 0$

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

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

$= - 27 - 27 + 30 + 24$

$= 54 - 54 = 0$

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

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

$= 64 - 48 - 40 + 24$

$= 84 - 84 = 0$

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

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