Calculator search results

Formula
Organize by substituting the expression
Answer
circle-check-icon
Expand the expression
Answer
circle-check-icon
expand-arrow-icon
expand-arrow-icon
expand-arrow-icon
expand-arrow-icon
expand-arrow-icon
Factorize the expression
Answer
circle-check-icon
expand-arrow-icon
$2 \left( x+1 \right) ^{ 2 } -3 \left( x+1 \right) +1$
$x \left ( 2 x + 1 \right )$
Substitute and transform it into the quadratic expression to arrange an equation
$\color{#FF6800}{ 2 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 1 }$
$ $ Substitute $ x + 1 $ with $ t$
$\color{#FF6800}{ 2 } \color{#FF6800}{ t } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ t } \color{#FF6800}{ + } \color{#FF6800}{ 1 }$
$\color{#FF6800}{ 2 } \color{#FF6800}{ t } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ t } \color{#FF6800}{ + } \color{#FF6800}{ 1 }$
$ $ Do factorization $ $
$\left ( \color{#FF6800}{ t } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ t } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right )$
$\left ( \color{#FF6800}{ t } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ t } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right )$
$ $ Substitute $ t $ with $ x + 1$
$\left ( \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \left ( \color{#FF6800}{ 2 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right )$
$\left ( \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \left ( 2 \left ( x + 1 \right ) - 1 \right )$
$ $ Get rid of unnecessary parentheses $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \left ( 2 \left ( x + 1 \right ) - 1 \right )$
$\left ( x \color{#FF6800}{ + } \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \left ( 2 \left ( x + 1 \right ) - 1 \right )$
$ $ Eliminate opponent number $ $
$x \left ( 2 \left ( x + 1 \right ) - 1 \right )$
$x \left ( \color{#FF6800}{ 2 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right )$
$ $ Do factorization $ $
$x \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right )$
$2 x ^ { 2 } + x$
Organize polynomials
$2 \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) ^ { \color{#FF6800}{ 2 } } - 3 \left ( x + 1 \right ) + 1$
$ $ Expand the binomial expression $ $
$2 \left ( \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) - 3 \left ( x + 1 \right ) + 1$
$\color{#FF6800}{ 2 } \left ( \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) - 3 \left ( x + 1 \right ) + 1$
$ $ Organize the expression with the distributive law $ $
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } + \color{#FF6800}{ 4 } \color{#FF6800}{ x } + \color{#FF6800}{ 2 } - 3 \left ( x + 1 \right ) + 1$
$2 x ^ { 2 } + 4 x + 2 \color{#FF6800}{ - } \color{#FF6800}{ 3 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) + 1$
$ $ Organize the expression with the distributive law $ $
$2 x ^ { 2 } + 4 x + 2 \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } + 1$
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ 1 }$
$ $ Organize the similar terms $ $
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \left ( \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) \color{#FF6800}{ x } \color{#FF6800}{ + } \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right )$
$2 x ^ { 2 } + \left ( \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) \color{#FF6800}{ x } + \left ( 2 - 3 + 1 \right )$
$ $ Organize the mononomial expression $ $
$2 x ^ { 2 } + \color{#FF6800}{ x } + \left ( 2 - 3 + 1 \right )$
$2 x ^ { 2 } + x + \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right )$
$ $ Arrange the constant term $ $
$2 x ^ { 2 } + x + \color{#FF6800}{ 0 }$
$2 x ^ { 2 } + x \color{#FF6800}{ + } \color{#FF6800}{ 0 }$
$ $ 0 does not change when you add or subtract $ $
$2 x ^ { 2 } + x$
$x \left ( 2 x + 1 \right )$
Arrange the expression in the form of factorization..
$\color{#FF6800}{ 2 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 1 }$
$ $ Expand the expression $ $
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ x }$
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ x }$
$ $ Bind the expressions with the common factor $ x$
$\color{#FF6800}{ x } \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right )$
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
Check solution
search-thumbnail-$s|ef\left(-1n$ $\left($ }\right)^{50}\ $\right)$ \ | | is\ equal\ to\ $S$ $s1S$ $S-1S$ $s2S$ $s50s$
7th-9th grade
Other
Check solution
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
Check solution
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
Check solution
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
Check solution
Have you found the solution you wanted?
Try again
Try more features at QANDA!
Search by problem image
Ask 1:1 question to TOP class teachers
AI recommend problems and video lecture
apple logogoogle play logo