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Formula
Calculate the value
Answer
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Use the synthetic division to find the quotient and the remainder
Answer
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$\left( -3a ^{ 5 } -2a ^{ 4 } +27a ^{ 2 } \right) \div \left( -3a ^{ 2 } \right)$
$\dfrac { 3 a ^ { 5 } + 2 a ^ { 4 } - 27 a ^ { 2 } } { 3 a ^ { 2 } }$
Arrange the rational expression
$\left ( \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ a } ^ { \color{#FF6800}{ 5 } } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ a } ^ { \color{#FF6800}{ 4 } } \color{#FF6800}{ + } \color{#FF6800}{ 27 } \color{#FF6800}{ a } ^ { \color{#FF6800}{ 2 } } \right ) \color{#FF6800}{ \div } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ a } ^ { \color{#FF6800}{ 2 } } \right )$
$ $ Calculate the multiplication expression $ $
$\color{#FF6800}{ \dfrac { 3 a ^ { 5 } + 2 a ^ { 4 } - 27 a ^ { 2 } } { 3 a ^ { 2 } } }$
$ $ Quotient $ : a ^ { 3 } + \dfrac { 2 a ^ { 2 } } { 3 } - 9 \\ $ Remainder $ : 0$
Use the synthetic division to find the quotient and the remainder
$\left ( \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ a } ^ { \color{#FF6800}{ 5 } } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ a } ^ { \color{#FF6800}{ 4 } } \color{#FF6800}{ + } \color{#FF6800}{ 27 } \color{#FF6800}{ a } ^ { \color{#FF6800}{ 2 } } \right ) \color{#FF6800}{ \div } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ a } ^ { \color{#FF6800}{ 2 } } \right )$
$ $ Divide $ - 3 a ^ { 5 } - 2 a ^ { 4 } + 27 a ^ { 2 } $ by $ - 3 a ^ { 2 } $ using the synthetic division $ $
$ $ Quotient $ : a ^ { 3 } + \dfrac { 2 a ^ { 2 } } { 3 } - 9 \\ $ Remainder $ : 0$
Solution search results
search-thumbnail-Select one expression equivalent to the perimeter of the triangle below. $2a^{5}-6$ $a^{5}+2a^{2}-3$ $a^{2}+a^{4}+4$ $A$ $2a^{10}-2a^{4}-5$ $B$ $3a^{5}+a^{4}+a^{2}-5$ - $c$ $2a^{5}+a^{4}-2a^{2}-5$ $0$ $\left(2a^{5}-6\right).\left(a^{5}+2a^{2}-3\right).\left(-a^{2}+α^{4}+4\right)$ $\left(2a^{10}-6\right)+\left(a^{5}+2a^{2}-3\right)+\left(-a^{2}+a^{4}+4\right)$ –
10th-13th grade
Other
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search-thumbnail-The is the statement of the Mean Value Theorem from your $te\times tb00k$ $Tneoren$ $m4.5$ $Ne8n$ Value Theorem Let f be continuous over $leclose$ interval $\left(a.b\right)aod$ $i$ $eo$ $xd$ $csnlc$ $\left(ab\right)$ $mmd$ exists at least one point cE $\left(a.b\right)$ $si1dhtba$ $f^{'}\left(c\right)=\dfrac {f\left(b\right)-f\left(a\right)} {b-a}$ What is the geometric interpretation of the conclusion of the theorem? O The tangent line to the graph of \(f\left(x\right)\) at $l\left(c1\right)$ is parallel to the secant line connecting \(\left(a,f\left(a\right)\right)\) and \(\left(b,f\left(b\right)\right)\). O The tangent line to the graph of $\left(f\left(lef\left(x\left(right\right)$ at \(c\) is the secant line connecting \(\left(a,f\left(a\right)\right)\) $ana$ \(\left(b,f\left(b\right)\right)\). O The tangent line to the graph $of$ $\left(f\left(leF\left(x\left(right\right)\left(\right)at1\left(c1\right)is$ perpendicular to the secant line connecting \(\left(a,f\left(a\right)\right)\) and A(\left(b,f\left(b\right)\right)\). O The tangent line to the graph of $\left(fleH\left(x\right)nght\right)\right)\right)$ $at$ $\left(c\right)\right)ishoizonta|$ $Tnere$ is more than one tangent line to the graph $0no$ $\left(flcf\left(x\right)night\right)\left($ $at1\left(c1\right)$
Calculus
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search-thumbnail-$5.$ Subtract $a^{4}+5a^{3}-3a^{2}$ from $7a^{5}-2a^{4}-a^{2}+3a$ $S0lution:$ $\left(7a^{5}-2a^{4}-a^{2}+3a\right)-\left(a^{4}+5a^{3}-3a^{2}\right)$ $=7a^{5}-2a^{4}-a^{2}+3a-a^{4}-5a^{3}+3a^{2}$ $=7a^{5}+\left(2a^{4}-a^{4}\right)-5a^{3}+\left(a^{2}+3a^{2}\right)+3a$ $=7a^{5}-3a^{4}-5a^{3}+2a^{2}+3a$ Subtracting vertically yields the same result. $7a^{5}$ $2a^{4}$ a? + $3a$ at + $5a^{3}$ $3a^{2}$ $7a^{5}$ - $3a^{4}$ $5a^{3}$ + $2a^{2}$ + $3a$
7th-9th grade
Other
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