You visited us 1 times! Enjoying our articles? Unlock Full Access!

Zeroes of a Polynomial

Let R1 and ...

Question

Let $R_{1}$ and $R_{2}$ are the remainders when polynomial $f (x) = 4 x^{3} + 3 x^{2} - 12 a x - 5$ and $g (x) = 2 x^{3} + a x^{2} - 6 x - 2$ are divided by $(x - 1)$ and $(x - 2)$ respectively. If $3 R_{1} + R_{2} - 28 = 0$ , the value of $a$ is

Open in App

Solution

$f (x) = 4 x^{3} + 3 x^{2} - 12 a x - 5$

$g (x) = 2 x^{3} + a x^{2} - 6 x - 2$

$f (x)$ when divided by $(x - 1)$ leaves remainder as $R_{1}$

$g (x)$ when divided by $(x - 2)$ leaves remainder as $R_{2}$

Therefore, according to Remainder theorem,

$f (1) = 4 {(1)}^{3} + 3 {(1)}^{2} - 12 a (1) - 5$

$= 4 + 3 - 12 a - 5$

$= 2 - 12 a = R_{1} ⟶ (1)$

Similarly,

$g (2) = 2 {(2)}^{3} + a {(2)}^{2} - 6 (2) - 2$

$= 16 + 4 a - 12 - 2$

$= 2 + 4 a = R_{2} ⟶ (2)$

Also, $3 R_{1} + R_{2} - 28 = 0$

$∴ 3 (2 - 12 a) + 2 + 4 a - 28 = 0$

$\Rightarrow 6 - 36 a + 2 + 4 a - 28 = 0$

$\Rightarrow 32 a = - 20$

$\Rightarrow a = \frac{- 5}{8}$

Suggest Corrections

Similar questions

2 x^{3} - 3 x^{2} + 1

x^{3} + 5 x - 3

r_{1}

r_{2}

(r_{1} r_{2} - r_{1} - r_{2})

a x^{3} + 3 x^{2} - 3

2 x^{3} - 5 x + a

R_{1},

R_{2}

2 R_{1} - R_{2} = 0.

The polynomial bx³+3x²-3 and 2x³-5x+b when divided by (x-4) leaves the remainders R₁ and R₂ respectively. Find the value of b if 2R₁-R₂=0

R_{1}

R_{2}

x^{3} + 2 x^{2} - 5 a x - 7

x^{3} + a x^{2} - 12 x + 6

x + 1

x - 2

2 R_{1} + R_{2} = 6

a

(2 x^{3} - 5 x^{2} + x + a)

(a x^{3} + 2 x^{3} - 3)

(x - 2)

R_{1}

R_{2}

^{'} a^{'}

R_{1} = R_{2}

Join BYJU'S Learning Program