Calculations
1)
For this question refer to the figure attached in which both synthetic & long divisions are performed.
The final answer is gained the same in both ways as 3 x 3 − 4 x 2 + 2 x − 1 ( x − 3 ) = ( 3 x 2 + 5 x + 17 ) + 50 ( x − 3 ) \qquad\qquad
\begin{aligned}
\small \frac{3x^3-4x^2+2x-1}{(x-3)}&=\small (3x^2 +5x+17)+\frac{50}{(x-3)}
\end{aligned} ( x − 3 ) 3 x 3 − 4 x 2 + 2 x − 1 = ( 3 x 2 + 5 x + 17 ) + ( x − 3 ) 50
2)
a.= 3 x 2 − 8 x + 4 − ( 6 x 2 + 7 x − 1 ) = 3 x 2 − 8 x + 4 − 6 x 2 − 7 x + 1 = − 3 x 2 − 15 x + 5 \qquad\qquad
\begin{aligned}
\small &=\small 3x^2-8x+4-(6x^2+7x-1)\\
&=\small 3x^2-8x+4-6x^2-7x+1\\
&=\small -3x^2-15x+5
\end{aligned} = 3 x 2 − 8 x + 4 − ( 6 x 2 + 7 x − 1 ) = 3 x 2 − 8 x + 4 − 6 x 2 − 7 x + 1 = − 3 x 2 − 15 x + 5
b.= 4 x 7 − 3 x 5 + 2 x + 4 + ( 12 x 5 − 3 x + 2 x 2 + 3 ) = 4 x 7 − 3 x 5 + 2 x + 4 + 12 x 5 − 3 x + 2 x 2 + 3 = 4 x 7 + 9 x 5 + 2 x 2 − x + 7 \qquad\qquad
\begin{aligned}
&=\small 4x^7-3x^5+2x+4+(12x^5-3x+2x^2+3)\\
&=\small 4x^7-3x^5+2x+4+12x^5-3x+2x^2+3\\
&=\small 4x^7+9x^5+2x^2-x+7
\end{aligned} = 4 x 7 − 3 x 5 + 2 x + 4 + ( 12 x 5 − 3 x + 2 x 2 + 3 ) = 4 x 7 − 3 x 5 + 2 x + 4 + 12 x 5 − 3 x + 2 x 2 + 3 = 4 x 7 + 9 x 5 + 2 x 2 − x + 7
c.= 3 x ( 5 x 2 − 2 x + 3 ) + 4 ( 5 x 2 − 2 x + 3 ) = 15 x 3 − 6 x 2 + 9 x + 20 x 2 − 8 x + 12 = 15 x 3 + 14 x 2 + x + 12 \qquad\qquad
\begin{aligned}
\small &=\small 3x(5x^2-2x+3)+4(5x^2-2x+3)\\
&=\small 15x^3-6x^2+9x+20x^2-8x+12\\
&=\small 15x^3+14x^2+x+12
\end{aligned} = 3 x ( 5 x 2 − 2 x + 3 ) + 4 ( 5 x 2 − 2 x + 3 ) = 15 x 3 − 6 x 2 + 9 x + 20 x 2 − 8 x + 12 = 15 x 3 + 14 x 2 + x + 12
d.= 3 + 4 i + ( 1 − 2 i ) = 3 + 4 i + 1 − 2 i = 4 + 2 i \qquad\qquad
\begin{aligned}
&=\small 3+4i+(1-2i)\\
&=\small 3+4i+1-2i\\
&=\small 4+2i
\end{aligned} = 3 + 4 i + ( 1 − 2 i ) = 3 + 4 i + 1 − 2 i = 4 + 2 i
e.= ( 2 − i ) ( 5 − 6 i ) = 10 − 12 i − 5 i + 6 i 2 = 10 − 17 i − 6 = 4 − 17 i \qquad\qquad
\begin{aligned}
&=\small (2-i)(5-6i)\\
&=\small 10-12i-5i+6i^2\\
&=\small 10-17i-6\\
&=\small 4-17i
\end{aligned} = ( 2 − i ) ( 5 − 6 i ) = 10 − 12 i − 5 i + 6 i 2 = 10 − 17 i − 6 = 4 − 17 i
f.= ( 3 − 7 i ) ( 1 − 2 i ) = 3 − 6 i − 7 i + 14 ( − 1 ) = − 11 − 13 i \qquad\qquad
\begin{aligned}
&=\small (3-7i)(1-2i)\\
&=\small 3-6i-7i+14(-1)\\
&=\small -11-13i
\end{aligned} = ( 3 − 7 i ) ( 1 − 2 i ) = 3 − 6 i − 7 i + 14 ( − 1 ) = − 11 − 13 i
3)
x 2 − ( − 81 ) = 0 x 2 − 81 i 2 = 0 ( x + 9 i ) ( x − 9 i ) = 0 x = ± 9 i \qquad\qquad
\begin{aligned}
\small x^2-(-81)&=\small 0\\
\small x^2-81i^2&=\small 0\\
\small (x+9i)(x-9i)&=\small 0\\
\small x&=\small \pm9i
\end{aligned} x 2 − ( − 81 ) x 2 − 81 i 2 ( x + 9 i ) ( x − 9 i ) x = 0 = 0 = 0 = ± 9 i
4)
7 x 3 = 3 ( 7 x 3 ) 3 = 3 3 7 x = 27 x = 27 7 \qquad\qquad
\begin{aligned}
\small \sqrt[3]{7x}&=\small 3\\
\small (\sqrt[3]{7x})^3&=\small 3^3\\
\small 7x&=\small 27\\
\small x&=\small \frac{27}{7}
\end{aligned} 3 7 x ( 3 7 x ) 3 7 x x = 3 = 3 3 = 27 = 7 27
Comments