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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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$\left( x-5 \right) ^{ 2 } -2 \left( x-5 \right) +1$
$\left ( x - 6 \right ) ^ { 2 }$
Substitute and transform it into the quadratic expression to arrange an equation
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 1 }$
$ $ Substitute $ x - 5 $ with $ t$
$\color{#FF6800}{ t } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ t } \color{#FF6800}{ + } \color{#FF6800}{ 1 }$
$\color{#FF6800}{ t } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ t } \color{#FF6800}{ + } \color{#FF6800}{ 1 }$
$ $ Do factorization $ $
$\left ( \color{#FF6800}{ t } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) ^ { \color{#FF6800}{ 2 } }$
$\left ( \color{#FF6800}{ t } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) ^ { \color{#FF6800}{ 2 } }$
$ $ Substitute $ t $ with $ x - 5$
$\left ( \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) ^ { \color{#FF6800}{ 2 } }$
$\left ( \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) ^ { 2 }$
$ $ Get rid of unnecessary parentheses $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) ^ { 2 }$
$\left ( x \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) ^ { 2 }$
$ $ Find the sum of the negative numbers $ $
$\left ( x \color{#FF6800}{ - } \color{#FF6800}{ 6 } \right ) ^ { 2 }$
$x ^ { 2 } - 12 x + 36$
Organize polynomials
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) ^ { \color{#FF6800}{ 2 } } - 2 \left ( x - 5 \right ) + 1$
$ $ Expand the binomial expression $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 10 } \color{#FF6800}{ x } + \color{#FF6800}{ 25 } - 2 \left ( x - 5 \right ) + 1$
$x ^ { 2 } - 10 x + 25 \color{#FF6800}{ - } \color{#FF6800}{ 2 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) + 1$
$ $ Organize the expression with the distributive law $ $
$x ^ { 2 } - 10 x + 25 \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } + \color{#FF6800}{ 10 } + 1$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 10 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 25 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 10 } \color{#FF6800}{ + } \color{#FF6800}{ 1 }$
$ $ Organize the similar terms $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 10 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) \color{#FF6800}{ x } \color{#FF6800}{ + } \left ( \color{#FF6800}{ 25 } \color{#FF6800}{ + } \color{#FF6800}{ 10 } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right )$
$x ^ { 2 } + \left ( \color{#FF6800}{ - } \color{#FF6800}{ 10 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) \color{#FF6800}{ x } + \left ( 25 + 10 + 1 \right )$
$ $ Arrange the constant term $ $
$x ^ { 2 } \color{#FF6800}{ - } \color{#FF6800}{ 12 } \color{#FF6800}{ x } + \left ( 25 + 10 + 1 \right )$
$x ^ { 2 } - 12 x + \left ( \color{#FF6800}{ 25 } \color{#FF6800}{ + } \color{#FF6800}{ 10 } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right )$
$ $ Arrange the constant term $ $
$x ^ { 2 } - 12 x + \color{#FF6800}{ 36 }$
$\left ( x - 6 \right ) ^ { 2 }$
Arrange the expression in the form of factorization..
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 1 }$
$ $ Expand the expression $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 12 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 36 }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 12 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 36 }$
$ $ Use the factoring formula, $ a^{2}-2ab + b^{2} = \left(a-b\right)^{2}$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 6 } \right ) ^ { \color{#FF6800}{ 2 } }$
Solution search results
search-thumbnail-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-$s|ef\left(-1n$ $\left($ }\right)^{50}\ $\right)$ \ | | is\ equal\ to\ $S$ $s1S$ $S-1S$ $s2S$ $s50s$
7th-9th 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-What is the inverse of $g\left(x\right)=\sqrt{x-1} $ $Og\left(x\right)=\sqrt{x+1} $ $+1$ $g^{'}\left(x\right)=5-$ $y_{9}$ $g\sqrt{ef\left(x\left(nght\right)} =\left(f_{tac\left(x+1\right)0sq\pi \left(x\right)}$ $Og^{'}\left(x\right)=x^{2}+1$ $○g\left(x\right)=x^{2}-1$ « Previous Ne
Other
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search-thumbnail-What is the inverse of $g\left(x\right)=\sqrt{x-1} $ $Og\left(x\right)=\sqrt{x+1} $ $+1$ $g^{'}\left(x\right)=5-$ $y_{9}$ $g\sqrt{ef\left(x\left(nght\right)} =\left(f_{tac\left(x+1\right)0sq\pi \left(x\right)}$ $Og^{'}\left(x\right)=x^{2}+1$ $○g\left(x\right)=x^{2}-1$ « Previous Ne
Other
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