Question

R1 and R2 are the remainder when f(x) equal 4x³+3x²-12ax -5and g(x) equal 2x³+ax²-6x-2 are divided by(x-1) and respectively. If3R1+R2-28equal0 find a

Answer 1

$\Longrightarrow{\boxed{\bold{Answer: \frac{-28}{35}}}}$

$\textbf{\underline{Step-by-step explanation :- }}$

➣ f(x) = 4x³ + 3x² - 12ax - 5

When x - 1 is a quotient then value of x = +1

f(1) = 4(1)³ + 3(1)² - 12a(1) - 5

R1 = 4 + 3 - 12a - 5

R1 = 2 - 12a

➣ g(x) = 2x³ + ax² - 6x - 2

g(1) = 2(1)³ + a(1)² - 6(1) - 2

R2 = 2 + a - 6 - 2

R2 = - 6 + a

➣ Now, 3R1 + R2 - 28 = 0

3( 2 - 12a ) + ( - 6 + a ) - 28 = 0

6 - 36a - 6 + a - 28 = 0

- 36a + a = 28

- 35a = 28

$\bold{a = \frac{28}{-35} = \frac{-28}{35}}$