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.
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Answered by
268
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☺!✌
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☺!✌
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Answered by
26
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
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