Question

The polynomial ax^3+3x^2-3 and 2x^2-5x+a when divided by (x-4) leave the remainder R1 and R2.find the value of a in each case. R1+R2=0, 2R1-R2=0.

Answer 1

Hello Mate!

Let the remainder be R1 and R2 as given :

R1 = ax^3 + 3x^2 - 3

Now, x - 4 => x = 4

R1 = a(4)^3 + 3(4)^2 - 3
R1 = 64a + 48 - 3
R1 = 64a + 45

R2 = 2x^2 - 5x + a
R2 = 2(4)^2 - 5(4) + a
R2 = 32 - 20 + a
R2 = 12 + a

R1 + R2 = 0

64a + 45 + 12 + a = 0
65a = - 57
a = - 57/65

2R1 - R2 = 0

2( 64a + 45 ) - 12 - a = 0
128a + 90 - 12 - a = 0
127a = - 78
a = - 78/127

But friend, same question I have also done but there is a slight difference in question. In my book there is 2x^3 - 5x + a instead of 2x^2 - 5x + a. So, I am giving pic of that too.

✴ If you had written 2x^2 instead of 2x^3 by mistake then refer to pic.
✴If not, then refer to text I wrote

Hope it helps☺!✌

Answer 2

Answer:  a=18/127         PLS MARK ME AS BRAINLIEST

sry friend in my book it is given as 2x^3 instead of 2x^2 . so i did for that only.

Step-by-step explanation:

x-4=0

so, x=4

• a×(4)^3+3×(4)^2-3
• a×64+3×16-3
• 64a+48 ⇒R1

• 2×x^3-5x+a
• 2×(4)^2-5×4+a
• 2×64-20+a
• 108+a⇒R2

2(R2) - R2

• 2(64a+45)-(108+a)=0

• 128a+90-108-a=0
• 128-18-a
• 127a=18
• a=18/127