Question

5. The polynomials 3x3 + ax2+ 3x + 5 and 4x3 + x2 – 2x + a leaves remainder R1 and R2, when divided by (x-2) respectively, if R1-R2=9, find the value of a .
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Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

p(x) = x4 – 2x3 + 3x2 – ax + 3a – 7

Divisor = x + 1

x + 1 = 0

x = -1

So, substituting the value of x = – 1 in p(x),

we get,

p(-1) = (-1)4 – 2(-1)3 + 3(-1)2 – a(-1) + 3a – 7.

19 = 1 + 2 + 3 + a + 3a – 7

19 = 6 – 7 + 4a

4a – 1 = 19

4a = 20

a = 5

Since, a = 5.

We get the polynomial,

p(x) = x4 – 2x3 + 3x2 – (5)x + 3(5) – 7

p(x) = x4 – 2x3 + 3x2 – 5x + 15 – 7

p(x) = x4 – 2x3 + 3x2 – 5x + 8

As per the question,

When the polynomial obtained is divided by (x + 2),

We get, x + 2 = 0

x = – 2

So, substituting the value of x = – 2 in p(x), we get,

p(-2) = (-2)4 – 2(-2)3 + 3(-2)2 – 5(-2) + 8

⇒ p(-2) = 16 + 16 + 12 + 10 + 8

⇒ p(-2) = 62 Therefore, the remainder = 62.