Question

the polynomials kx4+3x3+6 when divided by x-2 leaves remainder which is double the remainder left by the polynomial 2x3 + 17x3+k when divided by (x-2). find the value of k.
answer is k=5
please help me​

Answer 1

Appropriate Question:

The polynomials kx⁴ + 3x³ + 6 when divided by (x – 2) leaves remainder which is double the remainder left by the polynomial 2x³ + 17x + k when divided by (x – 2). find the value of k.

Answer:

5

Step-by-step explanation:

Firstly we'll find the expressions of the remainders of both polynomials and equate it according to the question in order to calculate the value of k.

Let,

⠀⠀⠀⠀★ p(x) = kx⁴ + 3x³ + 6

⠀⠀⠀⠀★ f(x) = 2x³ + 17x + k

Here, both divisor's is (x – 2). Let the divisor of p(x) be g(x)' and divisor of f(x) be g(x)".

Finding the zero of the g(x)' :

$\longrightarrow \sf{\quad { g(x)' = 0 }}$

Finding the zero of the polynomial g(x) means solving the equation g(x) = 0. Substituting the value of g(x).

$\longrightarrow \sf{\quad { x - 2 = 0 }}$

Transposing -2 from L.H.S to R.H.S, its sign will get changed.

$\longrightarrow \sf{\quad { x = 0 + 2}}$

Performing addition in R.H.S.

$\longrightarrow \quad { \textbf {\textsf{x = 2}}}$

Finding the remainder when p(x) is divided by g(x)' :

$\longrightarrow \sf{\quad { p(x) = kx^4 + 3x^3+6 }}$

Finding the value of p(2) to find the remainder when p(x) is divided by g(x)'.

$\longrightarrow \sf{\quad { p(2) = k(2)^4 + 3(2)^3+ 6 }}$

Writing the squares of the numbers.

$\longrightarrow \sf{\quad { p(2) = 16k + 3(8) + 6 }}$

Performing multiplication.

$\longrightarrow \sf{\quad { p(2) = 16k + 24 + 6 }}$

Performing addition.

$\longrightarrow \quad { \textbf {\textsf{p(2) = 16k + 30 }}}$

$\underline{ \qquad\qquad\qquad\qquad\qquad\qquad\qquad}$

Finding the zero of the g(x)" :

$\longrightarrow \sf{\quad { g(x)" = 0 }}$

Finding the zero of the polynomial g(x) means solving the equation g(x)" = 0. Substituting the value of g(x)".

$\longrightarrow \sf{\quad { x - 2 = 0 }}$

Transposing -2 from L.H.S to R.H.S, its sign will get changed.

$\longrightarrow \sf{\quad { x = 0 + 2}}$

Performing addition in R.H.S.

$\longrightarrow \quad { \textbf {\textsf{x = 2}}}$

Finding the remainder when f(x) is divided by g(x)" :

$\longrightarrow \sf{\quad { f(x) = 2x^3 + 17x + k}}$

Finding the value of f(2) to find the remainder when f(x) is divided by g(x)'.

$\longrightarrow \sf{\quad { f(2) = 2(2)^3 + 17(2) + k}}$

Writing the squares of the numbers and performing multiplication.

$\longrightarrow \sf{\quad { f(2) = 2(8) + 34 + k}}$

Performing multiplication and performing addition.

$\longrightarrow \sf{\quad { f(2) = 16 + 34 + k}}$

Performing addition.

$\longrightarrow \quad { \textbf {\textsf{f(2) = 50 + k }}}$

$\underline{ \qquad\qquad\qquad\qquad\qquad\qquad\qquad}$

― According to the question, p(x) when divided by g(x)' leaves remainder which is double the remainder left by f(x) when divided g(x)".

$\longrightarrow \sf{\quad { p(2) =2 \times f(2)}}$

Substitute the values.

$\longrightarrow \sf{\quad { 16k + 30= 2(50 + k)}}$

Performing multiplication in RHS.

$\longrightarrow \sf{\quad { 16k + 30 = 100 + 2k}}$

Transposing like terms.

$\longrightarrow \sf{\quad { 16k - 2k = 100 - 30}}$

Performing subtraction in LHS and RHS.

$\longrightarrow \sf{\quad { 14k = 70}}$

Transposing 14 from LHS to RHS.

$\longrightarrow \sf{\quad { k = \dfrac{70}{14} }}$

Dividing 70 by 14.

$\longrightarrow \quad \underline{\boxed { \textbf {\textsf{ k= 5 }}}}$

∴ The value of k is 5.

Answer 2

Step-by-step explanation:

Appropriate Question:

The polynomials kx⁴ + 3x³ + 6 when divided by (x – 2) leaves remainder which is double the remainder left by the polynomial 2x³ + 17x + k when divided by (x – 2). find the value of k.

Answer:

5

Step-by-step explanation:

Firstly we'll find the expressions of the remainders of both polynomials and equate it according to the question in order to calculate the value of k.

Let,

⠀⠀⠀⠀★ p(x) = kx⁴ + 3x³ + 6

⠀⠀⠀⠀★ f(x) = 2x³ + 17x + k

Here, both divisor's is (x – 2). Let the divisor of p(x) be g(x)' and divisor of f(x) be g(x)".

Finding the zero of the g(x)' :

\longrightarrow \sf{\quad { g(x)' = 0 }}⟶g(x)

′

=0

Finding the zero of the polynomial g(x) means solving the equation g(x) = 0. Substituting the value of g(x).

\longrightarrow \sf{\quad { x - 2 = 0 }}⟶x−2=0

Transposing -2 from L.H.S to R.H.S, its sign will get changed.

\longrightarrow \sf{\quad { x = 0 + 2}}⟶x=0+2

Performing addition in R.H.S.

\longrightarrow \quad { \textbf {\textsf{x = 2}}}⟶x = 2

Finding the remainder when p(x) is divided by g(x)' :

\longrightarrow \sf{\quad { p(x) = kx^4 + 3x^3+6 }}⟶p(x)=kx

4

+3x

3

+6

Finding the value of p(2) to find the remainder when p(x) is divided by g(x)'.

\longrightarrow \sf{\quad { p(2) = k(2)^4 + 3(2)^3+ 6 }}⟶p(2)=k(2)

4

+3(2)

3

+6

Writing the squares of the numbers.

\longrightarrow \sf{\quad { p(2) = 16k + 3(8) + 6 }}⟶p(2)=16k+3(8)+6

Performing multiplication.

\longrightarrow \sf{\quad { p(2) = 16k + 24 + 6 }}⟶p(2)=16k+24+6

Performing addition.

\longrightarrow \quad { \textbf {\textsf{p(2) = 16k + 30 }}}⟶p(2) = 16k + 30

\underline{ \qquad\qquad\qquad\qquad\qquad\qquad\qquad}

Finding the zero of the g(x)" :

\longrightarrow \sf{\quad { g(x)" = 0 }}⟶g(x)"=0

Finding the zero of the polynomial g(x) means solving the equation g(x)" = 0. Substituting the value of g(x)".

\longrightarrow \sf{\quad { x - 2 = 0 }}⟶x−2=0

Transposing -2 from L.H.S to R.H.S, its sign will get changed.

\longrightarrow \sf{\quad { x = 0 + 2}}⟶x=0+2

Performing addition in R.H.S.

\longrightarrow \quad { \textbf {\textsf{x = 2}}}⟶x = 2

Finding the remainder when f(x) is divided by g(x)" :

\longrightarrow \sf{\quad { f(x) = 2x^3 + 17x + k}}⟶f(x)=2x

3

+17x+k

Finding the value of f(2) to find the remainder when f(x) is divided by g(x)'.

\longrightarrow \sf{\quad { f(2) = 2(2)^3 + 17(2) + k}}⟶f(2)=2(2)

3

+17(2)+k

Writing the squares of the numbers and performing multiplication.

\longrightarrow \sf{\quad { f(2) = 2(8) + 34 + k}}⟶f(2)=2(8)+34+k

Performing multiplication and performing addition.

\longrightarrow \sf{\quad { f(2) = 16 + 34 + k}}⟶f(2)=16+34+k

Performing addition.

\longrightarrow \quad { \textbf {\textsf{f(2) = 50 + k }}}⟶f(2) = 50 + k

\underline{ \qquad\qquad\qquad\qquad\qquad\qquad\qquad}

― According to the question, p(x) when divided by g(x)' leaves remainder which is double the remainder left by f(x) when divided g(x)".

\longrightarrow \sf{\quad { p(2) =2 \times f(2)}}⟶p(2)=2×f(2)

Substitute the values.

\longrightarrow \sf{\quad { 16k + 30= 2(50 + k)}}⟶16k+30=2(50+k)

Performing multiplication in RHS.

\longrightarrow \sf{\quad { 16k + 30 = 100 + 2k}}⟶16k+30=100+2k

Transposing like terms.

\longrightarrow \sf{\quad { 16k - 2k = 100 - 30}}⟶16k−2k=100−30

Performing subtraction in LHS and RHS.

\longrightarrow \sf{\quad { 14k = 70}}⟶14k=70

Transposing 14 from LHS to RHS.

\longrightarrow \sf{\quad { k = \dfrac{70}{14} }}⟶k=

14

70

Dividing 70 by 14.

\longrightarrow \quad \underline{\boxed { \textbf {\textsf{ k= 5 }}}}⟶

k= 5

∴ The value of k is 5.