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What must be added to the polynomial f(x) = ${{x}^{4}}+2{{x}^{3}}-2{{x}^{2}}+x-1$ to get the resulting polynomial which is exactly divisible by ${{x}^{2}}+2x-3$?
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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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