Question

show that
a.) ( 2x-3) is a factor of x + 2x cube - 9x square + 12 .​

Answer 1

Step-by-step explanation:

=> 2x³ - 9x² + x + 12

2x-3 = 0

2x = 3

x = 3/2

=> 2(3/2)³ - 9(3/2)² + 3/2 + 12

=> 2(27/8) - 9(9/4) + 3/2 + 12

=> 27/4 - 81/4 + 3/2 + 12

=> 27-81+6+48/4

=> 81-81/4

=> 0/4

=> 0

Here, remainder = 0

Therefore, 2x-3 is the factor of given polynomial!

OR

2x-3=0

x= 3/2

Putting the value of x by 3/2

p(x)= 2x3-9x2+x+12

p(3/2)= 2 (3/2)^3-9(3/2)^2+(3/2)+12

p(3/2)= 2×27/8-9×9/4+3/2+12

p(3/2)= 27/4-81/4+3/2+12

p(3/2)= 27-81+6+48/4

p(3/2)= 81-81/4

p(3/2)= 0/4= 0

We got the remainder 0, therefore

2x-3 is a factor of 2x3-9x2+x+12.

Answer 2

2x-3 =0

2x =3

x = 3/2

p(x) = x + 2x^3 - 9x^2 + 12

p(3/2) = (3/2) +2(3/2)^3 -9(3/2)^2 +12

= (3/2) + (27/4) -(81/4) + 12

= (6/4) + (27/4) - (81/4) + ( 48/4)

= (81/4) - (81/4)

p(3/2) = 0

Therefore (2x-3) is a factor of x+2x^3 -9x^2 + 12.

Hence proved.