GSEB Class 10 Maths Solutions Chapter 2 Polynomials Exercise 2.3

Question 1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:
1. \( p(x) = x^3 - 3x^2 + 5x - 3 \); \( g(x) = x^2 - 2 \)
2. \( p(x) = x^4 - 3x^2 + 4x + 5 \), \( g(x) = x^2 + 1 - x \)
3. \( p(x) = x^4 - 5x + 6 \), \( g(x) = 2 - x^2 \)
Answer:
1. By using the division process on the given polynomials \( p(x) \) and \( g(x) \):
First, we divide \( x^3 - 3x^2 + 5x - 3 \) by \( x^2 - 2 \).
\[ \begin{array}{r} x - 3 \\ x^2 - 2 \overline{) x^3 - 3x^2 + 5x - 3} \\ - \underline{(x^3 \phantom{+0x^2} - 2x)} \\ \hline - 3x^2 + 7x - 3 \\ - \underline{(-3x^2 \phantom{+0x} + 6)} \\ \hline 7x - 9 \end{array} \]
The degree of \( 7x - 9 \) is 1, which is smaller than the degree of \( x^2 - 2 \) (which is 2).
\( \implies \) Quotient = \( x - 3 \)
Remainder = \( 7x - 9 \)
To check our work: Quotient \( \times \) Divisor + Remainder
\( = (x - 3) (x^2 - 2) + (7x - 9) \)
\( = x^3 - 2x - 3x^2 + 6 + 7x - 9 \)
\( = x^3 - 3x^2 + 5x - 3 \)
\( = \) Dividend.
2. For \( p(x) = x^4 - 3x^2 + 4x + 5 \) and \( g(x) = x^2 + 1 - x \):
The standard form for \( g(x) \) is \( x^2 - x + 1 \).
Next, by applying the division method to the polynomials \( p(x) \) and \( g(x) \):
\[ \begin{array}{r} x^2 + x - 3 \\ x^2 - x + 1 \overline{) x^4 - 3x^2 + 4x + 5} \\ - \underline{(x^4 - x^3 + x^2)} \\ \hline x^3 - 4x^2 + 4x + 5 \\ - \underline{(x^3 - x^2 + x)} \\ \hline - 3x^2 + 3x + 5 \\ - \underline{(-3x^2 + 3x - 3)} \\ \hline 8 \end{array} \]
The degree of the remainder 8 is 0, which is less than the degree of \( x^2 - x + 1 \) (which is 2).
So, Quotient = \( x^2 + x - 3 \)
Remainder = 8.
To check this: Quotient \( \times \) Divisor + Remainder
\( = (x^2 + x - 3) (x^2 - x + 1) + 8 \)
\( = x^4 - x^3 + x^2 + x^3 - x^2 + x - 3x^2 + 3x - 3 + 8 \)
\( = x^4 - 3x^2 + 4x + 5 \)
\( = \) Dividend.
3. For \( p(x) = x^4 - 5x + 6 \) and \( g(x) = 2 - x^2 \):
The standard form for \( g(x) \) is \( -x^2 + 2 \).
Then, by using the division process on the given polynomials \( p(x) \) and \( g(x) \):
\[ \begin{array}{r} -x^2 - 2 \\ -x^2 + 2 \overline{) x^4 - 5x + 6} \\ - \underline{(x^4 \phantom{+0x^3} - 2x^2)} \\ \hline 2x^2 - 5x + 6 \\ - \underline{(2x^2 \phantom{+0x} - 4)} \\ \hline - 5x + 10 \end{array} \]
The degree of \( -5x + 10 \) is 1, which is less than the degree of \( -x^2 + 2 \) (which is 2).
Therefore, the quotient is \( -x^2 - 2 \) and the remainder is \( -5x + 10 \).
To check: Quotient \( \times \) Divisor + Remainder
\( = (-x^2 - 2) \times (-x^2 + 2) + (-5x + 10) \)
\( = (-x^2)^2 - 4 - 5x + 10 \)
\( = x^4 - 5x + 6 \)
\( = \) Dividend.
In simple words: We performed polynomial long division for each pair of polynomials given. For each division, we identified the quotient (the result of the division) and the remainder (what is left over). We also showed how to verify these results by multiplying the quotient by the divisor and adding the remainder, which should give back the original dividend.

Exam Tip: Always arrange polynomials in standard descending order of powers before starting division to avoid errors, and remember to include zero coefficients for missing terms if needed.

Question 2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:
1. \( t^2 - 3 \), \( 2t^4 + 3t^3 - 2t^2 - 9t - 12 \)
2. \( x^2 + 3x + 1 \), \( 3x^4 + 5x^3 - 7x^2 + 2x + 2 \)
3. \( x^3 - 3x + 1 \), \( x^5 - 4x^3 + x^2 + 3x + 1 \)
Answer:
1. By using the division process for the given polynomials:
\[ \begin{array}{r} 2t^2 + 3t + 4 \\ t^2 - 3 \overline{) 2t^4 + 3t^3 - 2t^2 - 9t - 12} \\ - \underline{(2t^4 \phantom{+3t^3} - 6t^2)} \\ \hline 3t^3 + 4t^2 - 9t - 12 \\ - \underline{(3t^3 \phantom{+4t^2} - 9t)} \\ \hline 4t^2 \phantom{-9t} - 12 \\ - \underline{(4t^2 \phantom{-9t} - 12)} \\ \hline 0 \end{array} \]
Since the remainder is 0, the first polynomial is a factor of the second polynomial.
2. By using the division process for the given polynomials:
\[ \begin{array}{r} 3x^2 - 4x + 2 \\ x^2 + 3x + 1 \overline{) 3x^4 + 5x^3 - 7x^2 + 2x + 2} \\ - \underline{(3x^4 + 9x^3 + 3x^2)} \\ \hline - 4x^3 - 10x^2 + 2x + 2 \\ - \underline{(-4x^3 - 12x^2 - 4x)} \\ \hline 2x^2 + 6x + 2 \\ - \underline{(2x^2 + 6x + 2)} \\ \hline 0 \end{array} \]
As the remainder is 0, the first polynomial is a factor of the second polynomial.
3. By using the division process for the given polynomials:
\[ \begin{array}{r} x^2 - 1 \\ x^3 - 3x + 1 \overline{) x^5 - 4x^3 + x^2 + 3x + 1} \\ - \underline{(x^5 - 3x^3 + x^2)} \\ \hline - x^3 \phantom{+x^2} + 3x + 1 \\ - \underline{(-x^3 \phantom{+3x} + 3x - 1)} \\ \hline 2 \end{array} \]
As the remainder is 2 (which is not 0), the first polynomial is not a factor of the second polynomial.
In simple words: To see if one polynomial divides evenly into another (making it a factor), we used polynomial long division. If the remainder at the end of the division is zero, then it's a factor; otherwise, it's not.

Exam Tip: A polynomial \( g(x) \) is a factor of \( p(x) \) if and only if the remainder of the division \( p(x) \div g(x) \) is zero. This is a fundamental concept in polynomial algebra.

Question 3. Obtain all other zeroes of \( 3x^4 + 6x^3 – 2x^2 – 10x – 5 \), if two zeroes are \( \sqrt{\frac{5}{3}} \) and \( -\sqrt{\frac{5}{3}} \).
Answer:
Because two zeroes are \( \sqrt{\frac{5}{3}} \) and \( -\sqrt{\frac{5}{3}} \), it means that \( \left(x-\sqrt{\frac{5}{3}}\right)\left(x+\sqrt{\frac{5}{3}}\right) = x^{2}-\frac{5}{3} \) is a factor of the polynomial we have.
Next, we divide the polynomial by \( x^2 - \frac{5}{3} \).
\[ \begin{array}{r} 3x^2 + 6x + 3 \\ x^2 - \frac{5}{3} \overline{) 3x^4 + 6x^3 - 2x^2 - 10x - 5} \\ - \underline{(3x^4 \phantom{+6x^3} - 5x^2)} \\ \hline 6x^3 + 3x^2 - 10x - 5 \\ - \underline{(6x^3 \phantom{+3x^2} - 10x)} \\ \hline 3x^2 \phantom{-10x} - 5 \\ - \underline{(3x^2 \phantom{-10x} - 5)} \\ \hline 0 \end{array} \]
So, \( 3x^4 + 6x^3 - 2x^2 - 10x - 5 = \left(x^2 - \frac{5}{3}\right) (3x^2 + 6x + 3) \)
Now, we factorize the quadratic part: \( 3x^2 + 6x + 3 \)
\( = 3x^2 + 3x + 3x + 3 \)
\( = 3x(x + 1) + 3(x + 1) \)
\( = 3(x + 1) (x + 1) \)
Thus, its zeroes are -1 and -1.
\( \implies \) The remaining two zeroes of the given polynomial are -1 and -1.
In simple words: If we know some roots of a polynomial, we can find a factor. We divided the polynomial by this factor. The result was a simpler quadratic polynomial. By factoring that quadratic, we found the two other missing roots, which turned out to be -1 and -1.

Exam Tip: Remember that if \( a \) is a zero of a polynomial, then \( (x-a) \) is a factor. When two zeroes are \( \sqrt{k} \) and \( -\sqrt{k} \), their product of factors \( (x-\sqrt{k})(x+\sqrt{k}) = x^2 - k \) is a factor.

Question 4. On dividing \( x^3 – 3x^2 + x + 2 \) by a polynomial \( g(x) \), the quotient and remainder were \( x-2 \) and \( -2x + 4 \), respectively. Find \( g(x) \).
Answer:
Given:
\( p(x) = x^3 - 3x^2 + x + 2 \)
\( q(x) = x - 2 \)
\( r(x) = -2x + 4 \)
Using the division process for polynomials, we know the relationship:
\( p(x) = g(x) \times q(x) + r(x) \)
\( \implies x^3 - 3x^2 + x + 2 = (x - 2) g(x) + (-2x + 4) \)
\( \implies (x - 2) g(x) = x^3 - 3x^2 + x + 2 - (-2x + 4) \)
\( \implies (x - 2) g(x) = x^3 - 3x^2 + x + 2 + 2x - 4 \)
\( \implies (x - 2) g(x) = x^3 - 3x^2 + 3x - 2 \)
\( \therefore g(x) = (x^3 - 3x^2 + 3x - 2) \div (x - 2) \)
Now, we perform the division:
\[ \begin{array}{r} x^2 - x + 1 \\ x - 2 \overline{) x^3 - 3x^2 + 3x - 2} \\ - \underline{(x^3 - 2x^2)} \\ \hline - x^2 + 3x - 2 \\ - \underline{(-x^2 + 2x)} \\ \hline x - 2 \\ - \underline{(x - 2)} \\ \hline 0 \end{array} \]
Thus, \( g(x) = x^2 - x + 1 \).
In simple words: We used the basic division formula (Dividend = Divisor \( \times \) Quotient + Remainder) and rearranged it to find the unknown divisor, \( g(x) \). Then, we performed polynomial long division to calculate \( g(x) \) using the values we had.

Exam Tip: Remember the division algorithm formula \( p(x) = g(x) \cdot q(x) + r(x) \). This formula is key to solving problems where any of the polynomials \( p(x), g(x), q(x), \) or \( r(x) \) are unknown.

Question 5. Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
1. deg p(x) = deg q(x)
2. deg q(x) = deg r(x)
3. deg r(x) =0.
Answer:
1. deg p(x) = deg q(x)
Let's use \( p(x) = 8x^2 - 8x + 16 \) and \( g(x) = 8 \).
If we divide \( p(x) \) by \( g(x) \), then \( q(x) = x^2 - x + 2 \) and \( r(x) = 0 \).
Here, the degree of \( p(x) \) is 2 and the degree of \( q(x) \) is 2. So, deg \( p(x) = \) deg \( q(x) \).
We can see that \( p(x) = g(x) \times q(x) + r(x) \) holds true.
2. deg q(x) = deg r(x)
Let's consider \( p(x) = x^3 + x^2 + x + 1 \) and \( g(x) = x^2 - 1 \).
If we divide \( p(x) \) by \( g(x) \), then \( q(x) = x + 1 \) and \( r(x) = 2x + 2 \).
Here, the degree of \( q(x) \) is 1 and the degree of \( r(x) \) is 1. So, deg \( q(x) = \) deg \( r(x) \).
This example also satisfies \( p(x) = g(x) \times q(x) + r(x) \).
3. deg r(x) = 0.
Let's take \( p(x) = x^4 + x^3 + x^2 - 2x - 3 \) and \( g(x) = x^2 - 2 \).
If we divide \( p(x) \) by \( g(x) \), then \( q(x) = x^2 + x + 3 \) and \( r(x) = 3 \).
Here, the degree of \( r(x) \) is 0 (since 3 is a constant). So, deg \( r(x) = 0 \).
Here too, \( p(x) = g(x) \times q(x) + r(x) \) is satisfied.
In simple words: We gave examples for three different scenarios based on the degrees of the polynomials. For the first one, we chose polynomials where the main polynomial and the quotient had the same highest power. For the second, the quotient and the remainder had the same highest power. For the last one, the remainder was just a constant number, meaning its highest power was zero. All examples followed the polynomial division rule.

Exam Tip: Understanding the relationship between the degrees of the polynomials \( p(x), g(x), q(x), \) and \( r(x) \) (especially \( \text{deg } r(x) < \text{deg } g(x) \)) is crucial for constructing such examples and for solving division algorithm problems effectively.