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Organize by substituting the expression
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Expand the expression
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Factorize the expression
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$\left ( x - 2 \right ) \left ( x + 6 \right )$
Substitute and transform it into the quadratic expression to arrange an equation
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) \color{#FF6800}{ - } \color{#FF6800}{ 15 }$
$ $ Substitute $ x + 3 $ with $ t$
$\color{#FF6800}{ t } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ t } \color{#FF6800}{ - } \color{#FF6800}{ 15 }$
$\color{#FF6800}{ t } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ t } \color{#FF6800}{ - } \color{#FF6800}{ 15 }$
$ $ Do factorization $ $
$\left ( \color{#FF6800}{ t } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) \left ( \color{#FF6800}{ t } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right )$
$\left ( \color{#FF6800}{ t } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) \left ( \color{#FF6800}{ t } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right )$
$ $ Substitute $ t $ with $ x + 3$
$\left ( \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) \left ( \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right )$
$\left ( \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) \left ( \left ( x + 3 \right ) + 3 \right )$
$ $ Get rid of unnecessary parentheses $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) \left ( \left ( x + 3 \right ) + 3 \right )$
$\left ( x + 3 - 5 \right ) \left ( \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right )$
$ $ Get rid of unnecessary parentheses $ $
$\left ( x + 3 - 5 \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right )$
$\left ( x + \color{#FF6800}{ 3 } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) \left ( x + 3 + 3 \right )$
$ $ Subtract $ 5 $ from $ 3$
$\left ( x \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) \left ( x + 3 + 3 \right )$
$\left ( x - 2 \right ) \left ( x + \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right )$
$ $ Add $ 3 $ and $ 3$
$\left ( x - 2 \right ) \left ( x + \color{#FF6800}{ 6 } \right )$
$x ^ { 2 } + 4 x - 12$
Organize polynomials
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) ^ { \color{#FF6800}{ 2 } } - 2 \left ( x + 3 \right ) - 15$
$ $ Expand the binomial expression $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } + \color{#FF6800}{ 6 } \color{#FF6800}{ x } + \color{#FF6800}{ 9 } - 2 \left ( x + 3 \right ) - 15$
$x ^ { 2 } + 6 x + 9 \color{#FF6800}{ - } \color{#FF6800}{ 2 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) - 15$
$ $ Organize the expression with the distributive law $ $
$x ^ { 2 } + 6 x + 9 \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 6 } - 15$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 6 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 9 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ - } \color{#FF6800}{ 15 }$
$ $ Organize the similar terms $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \left ( \color{#FF6800}{ 6 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) \color{#FF6800}{ x } \color{#FF6800}{ + } \left ( \color{#FF6800}{ 9 } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ - } \color{#FF6800}{ 15 } \right )$
$x ^ { 2 } + \left ( \color{#FF6800}{ 6 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) \color{#FF6800}{ x } + \left ( 9 - 6 - 15 \right )$
$ $ Arrange the constant term $ $
$x ^ { 2 } + \color{#FF6800}{ 4 } \color{#FF6800}{ x } + \left ( 9 - 6 - 15 \right )$
$x ^ { 2 } + 4 x + \left ( \color{#FF6800}{ 9 } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ - } \color{#FF6800}{ 15 } \right )$
$ $ Arrange the constant term $ $
$x ^ { 2 } + 4 x \color{#FF6800}{ - } \color{#FF6800}{ 12 }$
$\left ( x - 2 \right ) \left ( x + 6 \right )$
Arrange the expression in the form of factorization..
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) \color{#FF6800}{ - } \color{#FF6800}{ 15 }$
$ $ Expand the expression $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 12 }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 12 }$
$ $ Use the factoring formula, $ x^{2} + \left(a+b\right)x + ab = \left(x+a\right)\left(x+b\right)$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 6 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right )$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 6 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right )$
$ $ Sort the factors $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 6 } \right )$
Solution search results
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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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-$The\ reciprocal\\ 0+11) \left(\frac{2}
$8$ $\left(1$ Point) $1\right)$ The\ reciprocal\\ $0+11\right)$ \left(\frac{2} $c\left(2\right)$ {5}\right)^0\ $\right)$ \ $1111s\right)$ $S$ $S1S$ $s3S$ $S4S$ $s2S$
7th-9th grade
Other
Search count: 4,895
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search-thumbnail-eft , t 4
eft , t $4$ sh, $A9$ woer $ca$ $3$
10th-13th grade
Other
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search-thumbnail-Given the set of ordered pairs ((-7.0),(-6,5),(-5,-3),(-1,2) (1,6),(2,-2) (5,3)(7,-8))
Given the set of ordered pairs $\left(\left(-7.0\right),\left(-6,5\right),\left(-5,-3\right),\left(-1,2\right)$ $\left(1,6\right),\left(2,-2\right)$ $\left(5,3\right)\left(7,-8\right)\right)$ Find f(7)fAleft(7\right) O a O b -8 6. $5$
7th-9th grade
Algebra
Search count: 119
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