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Hint: To find the algebraic expression to be added to ${{x}^{4}}+2{{x}^{3}}-2{{x}^{2}}+x-1$ so that it is divisible by ${{x}^{2}}+2x-3$, we perform long division method to divide ${{x}^{4}}+2{{x}^{3}}-2{{x}^{2}}+x-1$ and ${{x}^{2}}+2x-3$. Then the additive inverse of the remainder from this long division method would give us the expression to be added to ${{x}^{4}}+2{{x}^{3}}-2{{x}^{2}}+x-1$ so that it is perfectly divisible by ${{x}^{2}}+2x-3$

Complete step-by-step solution:

To explain this method, say we divide 234 by 10. We would then get 23 as the quotient and 4 as the remainder. Thus, we need to add the additive inverse of the remainder (that is -4) from 234 to make it divisible by 10 (Thus, 234 + (-4) = 230, this is now divisible by 10). Now, the long division method is shown below.

${{x}^{2}}$+ 1

${{x}^{2}}+2x-3$ $\left| \!{\overline {\,

{{x}^{4}}+2{{x}^{3}}-2{{x}^{2}}+x-1 \,}} \right. $

-$\left( {{x}^{4}}+2{{x}^{3}}-3{{x}^{2}} \right)$

\[\left| \!{\overline {\,

{{x}^{2}}+x-1\text{ } \,}} \right. \]

-$\left( {{x}^{2}}+2x-3 \right)$

\[\]

To understand this, we first write down the divisor (${{x}^{2}}+2x-3$) and dividend $({{x}^{4}}+2{{x}^{3}}-2{{x}^{2}}+x-1)$ as shown above. Next we start with the highest power of x and accordingly find the first term of quotient. Thus, in this case since ${{x}^{4}}$ was the highest power term in the dividend, we divide this by the highest term in the divisor (${{x}^{2}}$), thus we get, $\dfrac{{{x}^{4}}}{{{x}^{2}}}={{x}^{4-2}}={{x}^{2}}$.

Next, we multiply ${{x}^{2}}+2x-3$ and ${{x}^{2}}$(first quotient term) to get ${{x}^{4}}+2{{x}^{3}}-3{{x}^{2}}$. Then we subtract ${{x}^{4}}+2{{x}^{3}}-3{{x}^{2}}$ from ${{x}^{4}}+2{{x}^{3}}-2{{x}^{2}}+x-1$ (which is similar to the long division method). Finally, we obtain ${{x}^{2}}+x-1$ (from subtraction). We then apply the same technique again (but now, ${{x}^{2}}+x-1$ acts as the dividend). Thus, we divide the highest power of ${{x}^{2}}+x-1$ (that is ${{x}^{2}}$) by the highest power of the divisor. We get, $\dfrac{{{x}^{2}}}{{{x}^{2}}}$=1 (which is the next quotient term). We again multiply the divisor by this term (that is 1) and then perform subtraction to get (-x+2) as the remainder. We stop at this stage since now the highest power of (-x+2) is lower than the divisor itself.

Now, the additive inverse of the remainder is

=-(-x+2)

=x-2

Thus, we need to add (x-2) to ${{x}^{4}}+2{{x}^{3}}-2{{x}^{2}}+x-1$ , so that it is divisible by ${{x}^{2}}+2x-3$.

Note: An alternative way to solve this problem would be to assume the expression to be added as Ax+B (we don’t need to have ${{x}^{2}}$ and higher terms here since while assumption we assume the expression with highest order as 1 less than the highest order of the divisor. Thus, in this case since the highest power of divisor is ${{x}^{2}}$, thus, the highest power of expression to be added is x). Now,

$\begin{align}

& ={{x}^{2}}+2x-3 \\

& ={{x}^{2}}+3x-x-3 \\

& =x(x+3)-(x+3) \\

& =(x-1)(x+3) \\

\end{align}$

Let, g(x)=${{x}^{4}}+2{{x}^{3}}-2{{x}^{2}}+x-1+Ax+B$

Now, since 1 and 3 are the zeros of ${{x}^{2}}+2x-3$, thus they should also be zeros of g(x). Thus,

g(1) = 0 -- (1)

g(-3) = 0 -- (2)

From (1),

1+2-2+1-1+A+B=0

1+A+B=0

A+B=-1 -- (3)

From (2),

$-{{3}^{4}}+2{{(-3)}^{3}}-2{{(-3)}^{2}}+(-3)-1+A(-3)+B=0$

5-3A+B = 0 -- (4)

Subtracting (3) from (4), we get,

(5-3A+B) -(A+B) = 0-(-1)

5-4A=1

A=1 -- (5)

Now put this value of A in (3), we get,

1+B=-1

B=-2 -- (6)

Since, the expression to be added was Ax+B, thus, the expression to be added is (x-2).

Hence, we get the same answer as that in the solution.

Complete step-by-step solution:

To explain this method, say we divide 234 by 10. We would then get 23 as the quotient and 4 as the remainder. Thus, we need to add the additive inverse of the remainder (that is -4) from 234 to make it divisible by 10 (Thus, 234 + (-4) = 230, this is now divisible by 10). Now, the long division method is shown below.

${{x}^{2}}$+ 1

${{x}^{2}}+2x-3$ $\left| \!{\overline {\,

{{x}^{4}}+2{{x}^{3}}-2{{x}^{2}}+x-1 \,}} \right. $

-$\left( {{x}^{4}}+2{{x}^{3}}-3{{x}^{2}} \right)$

\[\left| \!{\overline {\,

{{x}^{2}}+x-1\text{ } \,}} \right. \]

-$\left( {{x}^{2}}+2x-3 \right)$

\[\]

To understand this, we first write down the divisor (${{x}^{2}}+2x-3$) and dividend $({{x}^{4}}+2{{x}^{3}}-2{{x}^{2}}+x-1)$ as shown above. Next we start with the highest power of x and accordingly find the first term of quotient. Thus, in this case since ${{x}^{4}}$ was the highest power term in the dividend, we divide this by the highest term in the divisor (${{x}^{2}}$), thus we get, $\dfrac{{{x}^{4}}}{{{x}^{2}}}={{x}^{4-2}}={{x}^{2}}$.

Next, we multiply ${{x}^{2}}+2x-3$ and ${{x}^{2}}$(first quotient term) to get ${{x}^{4}}+2{{x}^{3}}-3{{x}^{2}}$. Then we subtract ${{x}^{4}}+2{{x}^{3}}-3{{x}^{2}}$ from ${{x}^{4}}+2{{x}^{3}}-2{{x}^{2}}+x-1$ (which is similar to the long division method). Finally, we obtain ${{x}^{2}}+x-1$ (from subtraction). We then apply the same technique again (but now, ${{x}^{2}}+x-1$ acts as the dividend). Thus, we divide the highest power of ${{x}^{2}}+x-1$ (that is ${{x}^{2}}$) by the highest power of the divisor. We get, $\dfrac{{{x}^{2}}}{{{x}^{2}}}$=1 (which is the next quotient term). We again multiply the divisor by this term (that is 1) and then perform subtraction to get (-x+2) as the remainder. We stop at this stage since now the highest power of (-x+2) is lower than the divisor itself.

Now, the additive inverse of the remainder is

=-(-x+2)

=x-2

Thus, we need to add (x-2) to ${{x}^{4}}+2{{x}^{3}}-2{{x}^{2}}+x-1$ , so that it is divisible by ${{x}^{2}}+2x-3$.

Note: An alternative way to solve this problem would be to assume the expression to be added as Ax+B (we don’t need to have ${{x}^{2}}$ and higher terms here since while assumption we assume the expression with highest order as 1 less than the highest order of the divisor. Thus, in this case since the highest power of divisor is ${{x}^{2}}$, thus, the highest power of expression to be added is x). Now,

$\begin{align}

& ={{x}^{2}}+2x-3 \\

& ={{x}^{2}}+3x-x-3 \\

& =x(x+3)-(x+3) \\

& =(x-1)(x+3) \\

\end{align}$

Let, g(x)=${{x}^{4}}+2{{x}^{3}}-2{{x}^{2}}+x-1+Ax+B$

Now, since 1 and 3 are the zeros of ${{x}^{2}}+2x-3$, thus they should also be zeros of g(x). Thus,

g(1) = 0 -- (1)

g(-3) = 0 -- (2)

From (1),

1+2-2+1-1+A+B=0

1+A+B=0

A+B=-1 -- (3)

From (2),

$-{{3}^{4}}+2{{(-3)}^{3}}-2{{(-3)}^{2}}+(-3)-1+A(-3)+B=0$

5-3A+B = 0 -- (4)

Subtracting (3) from (4), we get,

(5-3A+B) -(A+B) = 0-(-1)

5-4A=1

A=1 -- (5)

Now put this value of A in (3), we get,

1+B=-1

B=-2 -- (6)

Since, the expression to be added was Ax+B, thus, the expression to be added is (x-2).

Hence, we get the same answer as that in the solution.

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