The polynomials bx² + 3 x²
3 x² - 3 and
2x - 52 tb leave req remainders
R1 & R2 respectively when divided
by a-u. If 2R1 - R2=0, Find the
value of b
please it's urgent. with full process
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Answer:
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☺!✌
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