Question

If $f\left( x \right)={{x}^{4}}+9{{x}^{3}}+35{{x}^{2}}-x+4$ , find $f\left( -5+4i \right)$

Answer 1

Hint: We need to find the value of the given function at $x=-5+4i$ . We start to solve the given question by finding out the quadratic equation in x. Then, we divide the function $f\left( x \right)$ with the quadratic equation to get the desired result.

Complete step by step answer:
We are given a function $f\left( x \right)$ and are asked to find the value of the function at $x=-5+4i$ . We will be solving the given question by finding the quadratic equation in x and then divide the function $f\left( x \right)$ with the quadratic equation.
According to our question,
$\Rightarrow x=-5+4i$
Moving the term -5 to the other side of the equation, we get,
$\Rightarrow x+5=4i$
Squaring the above equation on both sides, we get,
$\Rightarrow {{\left( x+5 \right)}^{2}}={{\left( 4i \right)}^{2}}$
Expanding the squares in the above equation, we get,
$\Rightarrow {{x}^{2}}+10x+25=16{{i}^{2}}$
From complex numbers, we know that the value of ${{i}^{2}}=-1$
Substituting the same in the above equation, we get,
$\Rightarrow {{x}^{2}}+10x+25=-16$
Moving the term -16 to the other side of the equation, we get,
$\Rightarrow {{x}^{2}}+10x+25+16=0$
Simplifying the above equation, we get,
$\Rightarrow {{x}^{2}}+10x+41=0$
Now, we need to divide If $f\left( x \right)={{x}^{4}}+9{{x}^{3}}+35{{x}^{2}}-x+4$ by ${{x}^{2}}+10x+41$
Here,
Dividend = ${{x}^{4}}+9{{x}^{3}}+35{{x}^{2}}-x+4$
Divisor = ${{x}^{2}}+10x+41$
We need to divide the first term of the dividend with the first term of the divisor.
$\Rightarrow \dfrac{{{x}^{4}}}{{{x}^{2}}}={{x}^{2}}$
Multiplying ${{x}^{2}}$ with the divisor,
$\Rightarrow {{x}^{2}}\left( {{x}^{2}}+10x+41 \right)$
Multiplying ${{x}^{2}}$ to each term of the expression, we get,
$\Rightarrow {{x}^{4}}+10{{x}^{3}}+41{{x}^{2}}$
Subtracting the dividend from the above expression,
$\Rightarrow \left( {{x}^{4}}+9{{x}^{3}}+35{{x}^{2}}-x+4 \right)-\left( {{x}^{4}}+10{{x}^{3}}+41{{x}^{2}} \right)$
Simplifying the above expression, we get,
$\Rightarrow -{{x}^{3}}-6{{x}^{2}}-x+4$
The above steps can be represented as follows,
${{x}^{2}}+10x+41\overset{{{x}^{2}}}{\overline{\left){\begin{align}
& {{x}^{4}}+9{{x}^{3}}+35{{x}^{2}}-x+4 \\
& {{x}^{4}}+10{{x}^{3}}+41{{x}^{2}} \\
& \underline{\left( - \right)\text{ }\left( - \right)\text{ }\left( - \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
& -{{x}^{3}}-6{{x}^{2}}-x+4 \\
\end{align}}\right.}}$
Now, we need to divide the first term of $-{{x}^{3}}-6{{x}^{2}}-x+4$ with the first term of the divisor.
$\Rightarrow \dfrac{-{{x}^{3}}}{{{x}^{2}}}=-x$
Multiplying $-x$ with the divisor,
$\Rightarrow -x\left( {{x}^{2}}+10x+41 \right)$
Multiplying $-x$ to each term of the expression, we get,
$\Rightarrow -{{x}^{3}}-10{{x}^{2}}-41x$
Subtracting $-{{x}^{3}}-6{{x}^{2}}-x$ and the above expression,
$\Rightarrow \left( -{{x}^{3}}-6{{x}^{2}}-x+4 \right)-\left( -{{x}^{3}}-10{{x}^{2}}-41x \right)$
Simplifying the above expression, we get,
$\Rightarrow 4{{x}^{2}}+40x+4$
The above steps can be represented as follows,
${{x}^{2}}+10x+41\overset{{{x}^{2}}-x}{\overline{\left){\begin{align}
& {{x}^{4}}+9{{x}^{3}}+35{{x}^{2}}-x+4 \\
& {{x}^{4}}+10{{x}^{3}}+41{{x}^{2}} \\
& \underline{\left( - \right)\text{ }\left( - \right)\text{ }\left( - \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
& -{{x}^{3}}-6{{x}^{2}}-x+4 \\
& \text{ }-{{x}^{3}}-10{{x}^{2}}-41x \\
& \,\underline{\text{ }\,\left( + \right)\,\,\left( + \right)\,\,\,\,\,\,\,\left( + \right)} \\
& 4{{x}^{2}}+40x+4 \\
\end{align}}\right.}}$
Now, we need to divide the first term of $4{{x}^{2}}+40x+4$ with the first term of the divisor.
$\Rightarrow \dfrac{4{{x}^{2}}}{{{x}^{2}}}=4$
Multiplying 4 with the divisor,
$\Rightarrow 4\left( {{x}^{2}}+10x+41 \right)$
Multiplying $4$ to each term of the expression, we get,
$\Rightarrow 4{{x}^{2}}+40x+164$
Subtracting $4{{x}^{2}}+40x+4$ from the above expression,
$\Rightarrow \left( 4{{x}^{2}}+40x+4 \right)-\left( 4{{x}^{2}}+40x+4 \right)$
Simplifying the above expression, we get,
$\Rightarrow -160$
The above steps can be represented as follows,
${{x}^{2}}+10x+41\overset{{{x}^{2}}-x+4}{\overline{\left){\begin{align}
& {{x}^{4}}+9{{x}^{3}}+35{{x}^{2}}-x+4 \\
& {{x}^{4}}+10{{x}^{3}}+41{{x}^{2}} \\
& \underline{\left( - \right)\text{ }\left( - \right)\text{ }\left( - \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
& -{{x}^{3}}-6{{x}^{2}}-x+4 \\
& \text{ }-{{x}^{3}}-10{{x}^{2}}-41x \\
& \,\underline{\text{ }\,\left( + \right)\,\,\left( + \right)\,\,\,\,\,\,\,\left( + \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
& 4{{x}^{2}}+40x+4 \\
& \text{ }4{{x}^{2}}+40x+164 \\
& \underline{\text{ }\left( - \right)\text{ }\left( - \right)\text{ }\,\,\,\left( - \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-160 \\
\end{align}}\right.}}$
The degree of the remainder polynomial $-160$ is 0.
The degree of the divisor polynomial $\left( {{x}^{2}}+10x+41 \right)$ is 2.
As the degree of the remainder polynomial is less than that of the divisor polynomial, the division cannot be further performed.
$\therefore$ The value of the function $f\left( x \right)={{x}^{4}}+9{{x}^{3}}+35{{x}^{2}}-x+4$ at $x=-5+4i$ is -160.

Note: The Long division method helps us to find the factors of the polynomial. The remainder and quotient of the division can be cross-checked using the formula,$\Rightarrow \dfrac{Dividend}{Divisor}=Quotient+\dfrac{\operatorname{Re}mainder}{Divisor}$