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Expand the expression
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Factorize the expression
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$\left( x ^{ 2 } +3x \right) ^{ 2 } -2x ^{ 2 } -6x-8$
$x ^ { 4 } + 6 x ^ { 3 } + 7 x ^ { 2 } - 6 x - 8$
Organize polynomials
$\left ( \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \right ) ^ { \color{#FF6800}{ 2 } } - 2 x ^ { 2 } - 6 x - 8$
$ $ Expand the binomial expression $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 4 } } + \color{#FF6800}{ 6 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 3 } } + \color{#FF6800}{ 9 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } - 2 x ^ { 2 } - 6 x - 8$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 4 } } \color{#FF6800}{ + } \color{#FF6800}{ 6 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 3 } } \color{#FF6800}{ + } \color{#FF6800}{ 9 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 8 }$
$ $ Organize the similar terms $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 4 } } \color{#FF6800}{ + } \color{#FF6800}{ 6 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 3 } } \color{#FF6800}{ + } \left ( \color{#FF6800}{ 9 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 8 }$
$x ^ { 4 } + 6 x ^ { 3 } + \left ( \color{#FF6800}{ 9 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } - 6 x - 8$
$ $ Arrange the constant term $ $
$x ^ { 4 } + 6 x ^ { 3 } + \color{#FF6800}{ 7 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } - 6 x - 8$
$\left ( x - 1 \right ) \left ( x + 1 \right ) \left ( x + 2 \right ) \left ( x + 4 \right )$
Arrange the expression in the form of factorization..
$\left ( \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 8 }$
$ $ Expand the expression $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 4 } } \color{#FF6800}{ + } \color{#FF6800}{ 6 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 3 } } \color{#FF6800}{ + } \color{#FF6800}{ 7 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 8 }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 4 } } \color{#FF6800}{ + } \color{#FF6800}{ 6 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 3 } } \color{#FF6800}{ + } \color{#FF6800}{ 7 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 8 }$
$ $ Do factorization $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \left ( \color{#FF6800}{ x } ^ { \color{#FF6800}{ 3 } } \color{#FF6800}{ + } \color{#FF6800}{ 7 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 14 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 8 } \right )$
$\left ( x - 1 \right ) \left ( \color{#FF6800}{ x } ^ { \color{#FF6800}{ 3 } } \color{#FF6800}{ + } \color{#FF6800}{ 7 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 14 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 8 } \right )$
$ $ Do factorization $ $
$\left ( x - 1 \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) \left ( \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 6 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 8 } \right )$
$\left ( x - 1 \right ) \left ( x + 1 \right ) \left ( \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 6 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 8 } \right )$
$ $ Use the factoring formula, $ x^{2} + \left(a+b\right)x + ab = \left(x+a\right)\left(x+b\right)$
$\left ( x - 1 \right ) \left ( x + 1 \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \right )$
$\left ( x - 1 \right ) \left ( x + 1 \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \right )$
$ $ Sort the factors $ $
$\left ( x - 1 \right ) \left ( x + 1 \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \right )$
Solution search results
search-thumbnail-Using the \emph{removal of first derivative}
Using the \emph{removal of first derivative} method, the differential equation \( \frac{d^{2}y} $\left(d\times n$ $\left(2\right)\right)+P|ffac\left(dy\right)\left(dx\right)+Qy=F$ $dx\right)+Qy=RN\right)$ is transformed as \). For, the differential equation \frac{d^{2}y} $\left(d^{n}\left(2\right)y\right)$ $dx$ $\left(2\right)+2x$ $\left(0C\left(dy\right)\left(dx\right)+\left(x$ $2+1\right)y=\times n3+3x\right)$ the value of $\left(11\right)$
Calculus
Search count: 3,465
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search-thumbnail-∠ F From the sum ot x^{A}(2)-8 with 3x-12 subtract the sum ofx-9
$∠$ $F$ From the sum $ot$ $x^{A}\left(2\right)-8$ with $3x-12$ subtract the sum $ofx-9$ with $3x-x^{A}\left(2\right)$ Multinlication takes place $\left(2\times A$ $\left(2\right)+5x+21\right)\left(x-$ $5\right)=$ #Please provide solution by using math formula.
10th-13th grade
Geometry
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search-thumbnail-s|ef(-1n }\right)^{50}\ ) \ is\ equal\ to\ S
$s|ef\left(-1n$ $\left($ }\right)^{50}\ $\right)$ \ | | is\ equal\ to\ $S$ $s1S$ $S-1S$ $s2S$ $s50s$
7th-9th grade
Other
Search count: 625
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search-thumbnail-Theis the statement of the Mean Value Theorem from your te\timestb00k
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
Search count: 3,278
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search-thumbnail-If the sum of two consecutive
If the sum of two consecutive numbers is $45$ and one number is $X$ .This statement in the form of equation $1s:$ $\left(1$ Point) $\right)$ $○5x+1$ $1eft\left(x+1$ $r1gnt\right)=45s$ $○sx+1ef\left(x+2$ $r1gnt\right)=145s$ $sx+1x=45s$
7th-9th grade
Algebra
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