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3 years ago

Based on the polynomial reminder theorem, what is the value of the function when x= -5

Mathematics

2 answers:

densk [106]3 years ago

5 0

Answer:

-15

Step-by-step explanation:

Given is a polynomial in x

$F (x)= x^4 + 12x^3 + 30x^2 - 12x + 70$

We have to find the remainder when the above polynomial is divided by x+5

Remainder theorem says that f(x) gives remainder R when divided by polynomial x-a means f(a) = R

Applying the above theorem we can say that value of the function when x =-5

= Remainder when f is divided by x+5

= F(-5)

Substitute the value of -5 in place of x

= (-5)^4 + 12(-5)^3 + 30(-5)^2 - 12(-5) + 70

= 625-1500+750+60+70

= 5

Hence answer is 5

kap26 [50]3 years ago

5 0

<u>Answer:</u>

$f(-5)=$ 5

<u>Step-by-step explanation:</u>

According to the polynomial remainder theorem, when a polynomial f(x) is divided by a linear polynomial (x - a), the remainder of that division will be equal to f(a).

Therefore, substituting the value of x as -5 in the given function to find its value:

$f(x)$ = $x^4 + 12x^3 + 30x^2 - 12x + 70$

$f(-5)$ = $(-5)^4 + 12(-5)^3 + 30(-5)^2 - 12(-5) + 70$

$f(-5)$ = $625+(-1500)+750-(-60)+70$

$f(-5)$ = $5$

Therefore, the value of the given function when x = -5 is 5.

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Read 2 more answers

Solve the system of equations by row-reduction. At each step, show clearly the symbol of the linear combinations that allow you

adell [148]

Answer:

1) The solution of the system is

$\left\begin{array}{ccc}x_1&=&5\\x_2&=&8\\x_3&=&-13\end{array}\right$

2) The solution of the system is

$\left\begin{array}{ccc}x_1&=&2\\x_2&=&-7\\x_3&=&-1\end{array}\right$

Step-by-step explanation:

1) To solve the system of equations

$\left\begin{array}{ccccccc}&3x_2&-5x_3&=&89\\6x_1&&+x_3&=&17\\x_1&-x_2&+8x_3&=&-107\end{array}\right$

using the row reduction method you must:

Step 1: Write the augmented matrix of the system

$\left[ \begin{array}{ccc|c} 0 & 3 & -5 & 89 \\\\ 6 & 0 & 1 & 17 \\\\ 1 & -1 & 8 & -107 \end{array} \right]$

Step 2: Swap rows 1 and 2

$\left[ \begin{array}{ccc|c} 6 & 0 & 1 & 17 \\\\ 0 & 3 & -5 & 89 \\\\ 1 & -1 & 8 & -107 \end{array} \right]$

Step 3: $\left(R_1=\frac{R_1}{6}\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 3 & -5 & 89 \\\\ 1 & -1 & 8 & -107 \end{array} \right]$

Step 4: $\left(R_3=R_3-R_1\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 3 & -5 & 89 \\\\ 0 & -1 & \frac{47}{6} & - \frac{659}{6} \end{array} \right]$

Step 5: $\left(R_2=\frac{R_2}{3}\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & -1 & \frac{47}{6} & - \frac{659}{6} \end{array} \right]$

Step 6: $\left(R_3=R_3+R_2\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & 0 & \frac{37}{6} & - \frac{481}{6} \end{array} \right]$

Step 7: $\left(R_3=\left(\frac{6}{37}\right)R_3\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & 0 & 1 & -13 \end{array} \right]$

Step 8: $\left(R_1=R_1-\left(\frac{1}{6}\right)R_3\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 5 \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & 0 & 1 & -13 \end{array} \right]$

Step 9: $\left(R_2=R_2+\left(\frac{5}{3}\right)R_3\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 5 \\\\ 0 & 1 & 0 & 8 \\\\ 0 & 0 & 1 & -13 \end{array} \right]$

Step 10: Rewrite the system using the row reduced matrix:

$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 5 \\\\ 0 & 1 & 0 & 8 \\\\ 0 & 0 & 1 & -13 \end{array} \right] \rightarrow \left\begin{array}{ccc}x_1&=&5\\x_2&=&8\\x_3&=&-13\end{array}\right$

2) To solve the system of equations

$\left\begin{array}{ccccccc}4x_1&-x_2&+3x_3&=&12\\2x_1&&+9x_3&=&-5\\x_1&+4x_2&+6x_3&=&-32\end{array}\right$

using the row reduction method you must:

Step 1:

$\left[ \begin{array}{ccc|c} 4 & -1 & 3 & 12 \\\\ 2 & 0 & 9 & -5 \\\\ 1 & 4 & 6 & -32 \end{array} \right]$

Step 2: $\left(R_1=\frac{R_1}{4}\right)$

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 2 & 0 & 9 & -5 \\\\ 1 & 4 & 6 & -32 \end{array} \right]$

Step 3: $\left(R_2=R_2-\left(2\right)R_1\right)$

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 0 & \frac{1}{2} & \frac{15}{2} & -11 \\\\ 1 & 4 & 6 & -32 \end{array} \right]$

Step 4: $\left(R_3=R_3-R_1\right)$

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 0 & \frac{1}{2} & \frac{15}{2} & -11 \\\\ 0 & \frac{17}{4} & \frac{21}{4} & -35 \end{array} \right]$

Step 5: $\left(R_2=\left(2\right)R_2\right)$

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 0 & 1 & 15 & -22 \\\\ 0 & \frac{17}{4} & \frac{21}{4} & -35 \end{array} \right]$

Step 6: $\left(R_1=R_1+\left(\frac{1}{4}\right)R_2\right)$

$\left[ \begin{array}{cccc} 1 & 0 & \frac{9}{2} & - \frac{5}{2} \\\\ 0 & 1 & 15 & -22 \\\\ 0 & \frac{17}{4} & \frac{21}{4} & -35 \end{array} \right]$

Step 7: $\left(R_3=R_3-\left(\frac{17}{4}\right)R_2\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & \frac{9}{2} & - \frac{5}{2} \\\\ 0 & 1 & 15 & -22 \\\\ 0 & 0 & - \frac{117}{2} & \frac{117}{2} \end{array} \right]$

Step 8: $\left(R_3=\left(- \frac{2}{117}\right)R_3\right)$

$\left[ \begin{array}{cccc} 1 & 0 & \frac{9}{2} & - \frac{5}{2} \\\\ 0 & 1 & 15 & -22 \\\\ 0 & 0 & 1 & -1 \end{array} \right]$

Step 9: $\left(R_1=R_1-\left(\frac{9}{2}\right)R_3\right)$

$\left[ \begin{array}{cccc} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 15 & -22 \\\\ 0 & 0 & 1 & -1 \end{array} \right]$

Step 10: $\left(R_2=R_2-\left(15\right)R_3\right)$

$\left[ \begin{array}{cccc} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 0 & -7 \\\\ 0 & 0 & 1 & -1 \end{array} \right]$

Step 11:

$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 0 & -7 \\\\ 0 & 0 & 1 & -1 \end{array} \right]\rightarrow \left\begin{array}{ccc}x_1&=&2\\x_2&=&-7\\x_3&=&-1\end{array}\right$