Question

help me with question 14' third part!
R1= 5a-6
R2=4a-10

please help me.. I'm posting it third time!​

Answer 1

I hope this might clear ur doubts.

Answer 2

Answer:

(i) a = (-4)

(ii) a = (16/9)

(iii) a = (11/7)

Step-by-step explanation:

We have,

x³ + 2x² - 5ax - 7 and x³ + ax² - 12x + 6 divided by (x + 1) and (x - 2) respectively.

Now,

Using Remainder theorem,

Let p(x) = x³ + 2x² - 5ax - 7

and g(x) = x + 1

Since,

p(x) is divided by g(x).

x + 1 = 0

x = -1

So,

p(-1) = (-1)³ + 2(-1)²- 5a(-1) - 7

R1 = -1 + 2(1) + 5a - 7

R1 = 5a - 8 + 2

R1 = 5a - 6

Similarly,

Using Remainder theorem,

Let f(y) = x³ + ax² - 12x + 6

g(y) = (x - 2)

So,

x - 2 = 0

x = 2

Then,

f(2) = 2³ + a(2)² - 12(2) + 6

R2 = 8 + 4a - 24 + 6

R2 = 4a - 24 + 14

R2 = 4a - 10

Now, according to the Question,

Case 1

R1 = R2

5a - 6 = 4a - 10

5a - 4a = 6 - 10

a = -4

Case 2

R1 + R2 = 0

(5a - 6) + (4a - 10) = 0

5a - 6 + 4a - 10 = 0

9a - 16 = 0

9a = 16

a = 16/9

Case 3

2R1 + R2 = 0

2(5a - 6) + (4a - 10) = 0

10a - 12 + 4a - 10 = 0

14a - 22 = 0

14a = 22

a = 22/14

a = 11/7

Hope it helped and believing you understood it........All the best