Math, asked by yuvesh, 1 day ago

Determine the exact value of k and m such for which the polynomial function p(x) = x^4 + x^3 + kx^2 + mx − 15 has all the following properties:
• the graph of p(x) crosses the x-axis at x = 1
• division by x + 1 gives a remainder of 2

Answers

Answered by mathdude500
4

\large\underline{\sf{Solution-}}

Given polynomial is

\rm :\longmapsto\:p(x) =  {x}^{4} +  {x}^{3} +  {kx}^{2} + mx - 15 \:  -  -  - (1)

Now, given that,

The graph of p(x) crosses the x-axis at x = 1.

It means, x = 1 is zero of p(x).

It means,

\rm :\longmapsto\:p(1) = 0

\rm :\longmapsto\:{1}^{4} +  {1}^{3} +  k + m - 15 \: =  \: 0

\rm :\longmapsto\:1 + 1 +  k + m - 15 \: =  \: 0

\rm :\longmapsto\:k + m - 13 \: =  \: 0

\rm :\longmapsto\:\boxed{\tt{ k + m  \: =  \: 13 }}-  -  -  - (2)

Also, given that

When p(x) is divided by x + 1, it gives a remainder of 2.

We know,

Remainder Theorem states that when a polynomial f(x) is divided by linear polynomial x - a, then remainder is f(a).

So,

\rm :\longmapsto\:p( - 1) = 2

\rm :\longmapsto\:{( - 1)}^{4} +  {( - 1)}^{3} +  {k( - 1)}^{2} + m( - 1) - 15 = 2

\rm :\longmapsto\:1 - 1 +k - m  - 15 = 2

\rm :\longmapsto\:k - m  - 15 = 2

\rm :\longmapsto\:k - m  = 2 + 15

\rm :\longmapsto\:\boxed{\tt{ k - m  = 17}} -  -  -  - (3)

On adding equation (2) and (3), we get

\rm :\longmapsto\:2k = 30

\rm\implies \:\boxed{\tt{  \: k \:  =  \: 15 \:  \: }} -  -  - (4)

On substituting the value of k in equation (3), we get

\rm :\longmapsto\:15 - m = 17

\rm :\longmapsto\: - m = 17 - 15

\rm :\longmapsto\: - m = 2

\rm\implies \:\boxed{\tt{  \: m \:  =  \:  -  \: 2 \:  \: }}

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More to know :-

ALGEBRAIC IDENTITIES

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

a² - b² = (a + b)(a - b)

(a + b)² = (a - b)² + 4ab

(a - b)² = (a + b)² - 4ab

(a + b)² + (a - b)² = 2(a² + b²)

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - b³ - 3ab(a - b)

Answered by EmperorSoul
1

\large\underline{\sf{Solution-}}

Given polynomial is

\rm :\longmapsto\:p(x) =  {x}^{4} +  {x}^{3} +  {kx}^{2} + mx - 15 \:  -  -  - (1)

Now, given that,

The graph of p(x) crosses the x-axis at x = 1.

It means, x = 1 is zero of p(x).

It means,

\rm :\longmapsto\:p(1) = 0

\rm :\longmapsto\:{1}^{4} +  {1}^{3} +  k + m - 15 \: =  \: 0

\rm :\longmapsto\:1 + 1 +  k + m - 15 \: =  \: 0

\rm :\longmapsto\:k + m - 13 \: =  \: 0

\rm :\longmapsto\:\boxed{\tt{ k + m  \: =  \: 13 }}-  -  -  - (2)

Also, given that

When p(x) is divided by x + 1, it gives a remainder of 2.

We know,

Remainder Theorem states that when a polynomial f(x) is divided by linear polynomial x - a, then remainder is f(a).

So,

\rm :\longmapsto\:p( - 1) = 2

\rm :\longmapsto\:{( - 1)}^{4} +  {( - 1)}^{3} +  {k( - 1)}^{2} + m( - 1) - 15 = 2

\rm :\longmapsto\:1 - 1 +k - m  - 15 = 2

\rm :\longmapsto\:k - m  - 15 = 2

\rm :\longmapsto\:k - m  = 2 + 15

\rm :\longmapsto\:\boxed{\tt{ k - m  = 17}} -  -  -  - (3)

On adding equation (2) and (3), we get

\rm :\longmapsto\:2k = 30

\rm\implies \:\boxed{\tt{  \: k \:  =  \: 15 \:  \: }} -  -  - (4)

On substituting the value of k in equation (3), we get

\rm :\longmapsto\:15 - m = 17

\rm :\longmapsto\: - m = 17 - 15

\rm :\longmapsto\: - m = 2

\rm\implies \:\boxed{\tt{  \: m \:  =  \:  -  \: 2 \:  \: }}

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

More to know :-

ALGEBRAIC IDENTITIES

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

a² - b² = (a + b)(a - b)

(a + b)² = (a - b)² + 4ab

(a - b)² = (a + b)² - 4ab

(a + b)² + (a - b)² = 2(a² + b²)

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - b³ - 3ab(a - b)

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