
How do you factor ${{x}^{10}}-{{y}^{2}}$ ?
Answer
558k+ views
Hint: We can factor the equation given in the question by applying the algebraic formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ we can assume ${{x}^{10}}$ is equal to ${{a}^{2}}$ and ${{y}^{2}}$ is equal to ${{b}^{2}}$ then apply the formula to factor the equation.
Complete step by step answer:
The given equation in the question is ${{x}^{10}}-{{y}^{2}}$
We the algebraic formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$
We can apply it to factor ${{x}^{10}}-{{y}^{2}}$
Let's assume ${{x}^{10}}$ is equal to ${{a}^{2}}$ and ${{y}^{2}}$ is equal to ${{b}^{2}}$ so the value of a is ${{x}^{5}}$ and value of b is y
We can write ${{x}^{10}}-{{y}^{2}}={{\left( {{x}^{5}} \right)}^{2}}-{{\left( y \right)}^{2}}$
Applying the formula
${{\left( {{x}^{5}} \right)}^{2}}-{{\left( y \right)}^{2}}=\left( {{x}^{5}}+y \right)\left( {{x}^{5}}-y \right)$
The factorization form of ${{x}^{10}}-{{y}^{2}}$ is $\left( {{x}^{5}}+y \right)\left( {{x}^{5}}-y \right)$
Note: We can solve the equation by another method . In the given equation ${{x}^{10}}-{{y}^{2}}$ let’s add and subtract the term $y{{x}^{5}}$ so ${{x}^{10}}-{{y}^{2}}$ = ${{x}^{10}}-{{x}^{5}}y+{{x}^{5}}y-{{y}^{2}}$
Now taking ${{x}^{5}}$ common in the first half of the equation and y common in second half of the equation we get
$\Rightarrow {{x}^{10}}-{{x}^{5}}y+{{x}^{5}}y-{{y}^{2}}={{x}^{5}}\left( {{x}^{5}}-y \right)+y\left( {{x}^{5}}-y \right)$
Now taking ${{x}^{5}}-y$ common from the equation we get
$\Rightarrow {{x}^{10}}-{{x}^{5}}y+{{x}^{5}}y-{{y}^{2}}=\left( {{x}^{5}}+y \right)\left( {{x}^{5}}-y \right)$
We can see the factor form of the equation is $\left( {{x}^{5}}+y \right)\left( {{x}^{5}}-y \right)$
Whenever we have a polynomial we equation we have to factor we can use the remainder theorem that if any number a satisfies $f\left( x \right)=0$ then x – a is a factor of the polynomial f(x)
When f(x) is divided by x – a then the reminder is f(a) . We can then divide f(x) by x – a to find the rest of the term of the equation and try to factorize if possible. Always remember some standard formula which will be helpful to solve many problems.
${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$
${{\left( x+y \right)}^{n}}=\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}{{x}^{n-r}}}{{y}^{r}}$
$\left( {{a}^{n}}-{{b}^{n}} \right)=\left( a-b \right)\sum\limits_{r=0}^{n-1}{{{a}^{n-1-r}}{{b}^{r}}}$ where n is a positive integer
Complete step by step answer:
The given equation in the question is ${{x}^{10}}-{{y}^{2}}$
We the algebraic formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$
We can apply it to factor ${{x}^{10}}-{{y}^{2}}$
Let's assume ${{x}^{10}}$ is equal to ${{a}^{2}}$ and ${{y}^{2}}$ is equal to ${{b}^{2}}$ so the value of a is ${{x}^{5}}$ and value of b is y
We can write ${{x}^{10}}-{{y}^{2}}={{\left( {{x}^{5}} \right)}^{2}}-{{\left( y \right)}^{2}}$
Applying the formula
${{\left( {{x}^{5}} \right)}^{2}}-{{\left( y \right)}^{2}}=\left( {{x}^{5}}+y \right)\left( {{x}^{5}}-y \right)$
The factorization form of ${{x}^{10}}-{{y}^{2}}$ is $\left( {{x}^{5}}+y \right)\left( {{x}^{5}}-y \right)$
Note: We can solve the equation by another method . In the given equation ${{x}^{10}}-{{y}^{2}}$ let’s add and subtract the term $y{{x}^{5}}$ so ${{x}^{10}}-{{y}^{2}}$ = ${{x}^{10}}-{{x}^{5}}y+{{x}^{5}}y-{{y}^{2}}$
Now taking ${{x}^{5}}$ common in the first half of the equation and y common in second half of the equation we get
$\Rightarrow {{x}^{10}}-{{x}^{5}}y+{{x}^{5}}y-{{y}^{2}}={{x}^{5}}\left( {{x}^{5}}-y \right)+y\left( {{x}^{5}}-y \right)$
Now taking ${{x}^{5}}-y$ common from the equation we get
$\Rightarrow {{x}^{10}}-{{x}^{5}}y+{{x}^{5}}y-{{y}^{2}}=\left( {{x}^{5}}+y \right)\left( {{x}^{5}}-y \right)$
We can see the factor form of the equation is $\left( {{x}^{5}}+y \right)\left( {{x}^{5}}-y \right)$
Whenever we have a polynomial we equation we have to factor we can use the remainder theorem that if any number a satisfies $f\left( x \right)=0$ then x – a is a factor of the polynomial f(x)
When f(x) is divided by x – a then the reminder is f(a) . We can then divide f(x) by x – a to find the rest of the term of the equation and try to factorize if possible. Always remember some standard formula which will be helpful to solve many problems.
${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$
${{\left( x+y \right)}^{n}}=\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}{{x}^{n-r}}}{{y}^{r}}$
$\left( {{a}^{n}}-{{b}^{n}} \right)=\left( a-b \right)\sum\limits_{r=0}^{n-1}{{{a}^{n-1-r}}{{b}^{r}}}$ where n is a positive integer
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