CBSE Class 10 Mathematics Polynomials Worksheet Set 02

Polynomials MCQ Questions with Answers Class 10 

Polynomials MCQ Questions with Answers Class 10 

Question 1. Find the zeros of each of the following polynomials and Verify the relationship between the zeros and their coefficients: i) \( 7x^2-25x-12 \) ii) \( 4x^2-17x-21 \) iii) \( \sqrt{3} x^2 + 10x + 7\sqrt{3} \)
Answer:

(i) For \( 7x^2 - 25x - 12 \):
We factorise by splitting the middle term:
\( 7x^2 - 28x + 3x - 12 = 0 \)
\( \implies 7x(x - 4) + 3(x - 4) = 0 \)
\( \implies (7x + 3)(x - 4) = 0 \)
The zeroes are \( \alpha = 4 \) and \( \beta = -\frac{3}{7} \).
Verification:
Sum of zeroes: \( \alpha + \beta = 4 + \left(-\frac{3}{7}\right) = \frac{25}{7} \). Also, \( -\frac{b}{a} = -\frac{-25}{7} = \frac{25}{7} \). (Verified)
Product of zeroes: \( \alpha\beta = 4 \cdot \left(-\frac{3}{7}\right) = -\frac{12}{7} \). Also, \( \frac{c}{a} = -\frac{12}{7} \). (Verified)

(ii) For \( 4x^2 - 17x - 21 \):
We factorise by splitting the middle term:
\( 4x^2 + 4x - 21x - 21 = 0 \)
\( \implies 4x(x + 1) - 21(x + 1) = 0 \)
\( \implies (4x - 21)(x + 1) = 0 \)
The zeroes are \( \alpha = \frac{21}{4} \) and \( \beta = -1 \).
Verification:
Sum of zeroes: \( \alpha + \beta = \frac{21}{4} - 1 = \frac{17}{4} \). Also, \( -\frac{b}{a} = -\frac{-17}{4} = \frac{17}{4} \). (Verified)
Product of zeroes: \( \alpha\beta = \frac{21}{4} \cdot (-1) = -\frac{21}{4} \). Also, \( \frac{c}{a} = -\frac{21}{4} \). (Verified)

(iii) For \( \sqrt{3} x^2 + 10x + 7\sqrt{3} \):
We factorise by splitting the middle term:
\( \sqrt{3}x^2 + 3x + 7x + 7\sqrt{3} = 0 \)
\( \implies \sqrt{3}x(x + \sqrt{3}) + 7(x + \sqrt{3}) = 0 \)
\( \implies (\sqrt{3}x + 7)(x + \sqrt{3}) = 0 \)
The zeroes are \( \alpha = -\sqrt{3} \) and \( \beta = -\frac{7}{\sqrt{3}} \).
Verification:
Sum of zeroes: \( \alpha + \beta = -\sqrt{3} + \left(-\frac{7}{\sqrt{3}}\right) = \frac{-3-7}{\sqrt{3}} = -\frac{10}{\sqrt{3}} \). Also, \( -\frac{b}{a} = -\frac{10}{\sqrt{3}} \). (Verified)
Product of zeroes: \( \alpha\beta = (-\sqrt{3})\left(-\frac{7}{\sqrt{3}}\right) = 7 \). Also, \( \frac{c}{a} = \frac{7\sqrt{3}}{\sqrt{3}} = 7 \). (Verified)
In simple words: Find the roots of each equation by factoring the middle term. Afterward, check that the sum of these roots is equal to \( -b/a \) and their product is equal to \( c/a \) to confirm the coefficients match the roots.

Exam Tip: Be sure to write out both the calculation of the roots and the verification equations. Clearly listing both helps prevent minor errors.

Question 2. In the adj. figure, the graph of a quadratic polynomial f(x) is given Find the zeros of f(x).
Answer:
By analyzing the given graph of the quadratic polynomial \( f(x) \), we can observe that the parabola intersects the horizontal x-axis at two distinct points, \( A(2, 0) \) and \( B(6, 0) \).
Since the zeroes of any polynomial are represented by the x-coordinates of the points where its graph intersects the x-axis, the zeroes of \( f(x) \) are \( x = 2 \) and \( x = 6 \).
In simple words: Look at where the curve crosses the horizontal line. It passes through the points at 2 and 6, so those two numbers are the zeroes of the polynomial.

Exam Tip: In graphical questions, the zeroes of the function are simply the x-coordinates of the points where the graph crosses the x-axis. No complex calculation is necessary.

Question 3. Find the coordinates of the points where the graph of the polynomial \( y = x^2+x-6 \) intersects x-axis.
Answer: The graph of the given polynomial intersects the x-axis at points where the y-value is exactly 0.
Setting \( y = 0 \):
\( x^2 + x - 6 = 0 \)
Factoring the quadratic equation:
\( (x + 3)(x - 2) = 0 \)
\( \implies x = -3 \) or \( x = 2 \).
So, the graph intersects the x-axis at the points \( (-3, 0) \) and \( (2, 0) \).
In simple words: Setting the equation to zero and factoring it gives the roots \( x = -3 \) and \( x = 2 \). Since these points lie on the horizontal axis, their coordinates are \( (-3, 0) \) and \( (2, 0) \).

Exam Tip: Always write your final answers as coordinates of the form \( (x, y) \), where the y-value is 0, since the question asks for coordinates rather than just the roots.

Question 4. Find the zeros of the cubic polynomial \( x^3+2x^2-x-2 \).
Answer: Let \( p(x) = x^3 + 2x^2 - x - 2 \). We can find its zeroes by grouping the terms:
\( x^2(x + 2) - 1(x + 2) = 0 \)
\( \implies (x^2 - 1)(x + 2) = 0 \)
\( \implies (x - 1)(x + 1)(x + 2) = 0 \)
Setting each factor to zero, we get:
\( x = 1 \), \( x = -1 \), or \( x = -2 \).
Thus, the zeroes of the polynomial are \( 1 \), \( -1 \), and \( -2 \).
In simple words: Grouping terms allows us to factorise the cubic equation into three linear parts. Solving these parts gives us the roots: \( 1 \), \( -1 \), and \( -2 \).

Exam Tip: Grouping is the fastest way to solve cubics when the coefficients are symmetric. If grouping fails, use the rational root theorem to find one zero first.

Question 5. Find a cubic polynomial whose zeros are 1, -2 and 3
Answer: Let the zeroes be \( \alpha = 1 \), \( \beta = -2 \), and \( \gamma = 3 \).
A cubic polynomial with these zeroes can be written as:
\( p(x) = (x - \alpha)(x - \beta)(x - \gamma) \)
\( \implies p(x) = (x - 1)(x + 2)(x - 3) \)
Multiplying the first two binomials:
\( p(x) = (x^2 + x - 2)(x - 3) \)
Multiplying the remaining terms:
\( p(x) = x(x^2 + x - 2) - 3(x^2 + x - 2) \)
\( \implies p(x) = x^3 + x^2 - 2x - 3x^2 - 3x + 6 \)
\( \implies p(x) = x^3 - 2x^2 - 5x + 6 \).
Thus, the required cubic polynomial is \( x^3 - 2x^2 - 5x + 6 \).
In simple words: We place the three roots into binomial factors as \( (x-1)(x+2)(x-3) \). Expanding and simplifying these brackets gives us the final cubic equation.

Exam Tip: You can also find the coefficients using symmetric sum relationships: \( -(\alpha+\beta+\gamma) \), \( (\alpha\beta+\beta\gamma+\gamma\alpha) \), and \( -(\alpha\beta\gamma) \).

Question 6. Find a quadratic polynomial whose zeros are \( 5+\sqrt{2} \) and \( 5-\sqrt{2} \)
Answer: Let the zeroes be \( \alpha = 5 + \sqrt{2} \) and \( \beta = 5 - \sqrt{2} \).
Sum of the zeroes, \( S = \alpha + \beta = (5 + \sqrt{2}) + (5 - \sqrt{2}) = 10 \)
Product of the zeroes, \( P = \alpha\beta = (5 + \sqrt{2})(5 - \sqrt{2}) = 5^2 - (\sqrt{2})^2 = 25 - 2 = 23 \)

The general formula for a quadratic polynomial is:
\( p(x) = x^2 - Sx + P \)
Substituting our calculated sum and product:
\( p(x) = x^2 - 10x + 23 \).
Thus, the quadratic polynomial is \( x^2 - 10x + 23 \).
In simple words: We add the roots to get 10 and multiply them to get 23. Putting these values into the standard template gives us \( x^2 - 10x + 23 \).

Exam Tip: Remember to use the difference of squares identity \( (a-b)(a+b) = a^2 - b^2 \) when multiplying conjugates to find the product of roots quickly.

Question 7. Find a quadratic polynomial, the sum of whose zeros is 8 and their product is 12. Hence find the zeros of the polynomial.
Answer: We are given:
Sum of zeroes, \( S = 8 \)
Product of zeroes, \( P = 12 \)

The formula for a quadratic polynomial is:
\( p(x) = x^2 - Sx + P \)
Substituting our sum and product values:
\( p(x) = x^2 - 8x + 12 \).

Now, we solve for the zeroes by factoring the polynomial:
\( x^2 - 8x + 12 = 0 \)
\( \implies (x - 2)(x - 6) = 0 \)
\( \implies x = 2 \) or \( x = 6 \).
So, the zeroes are \( 2 \) and \( 6 \).
In simple words: Putting the sum (8) and product (12) into our formula gives \( x^2 - 8x + 12 \). Factoring this equation yields the individual roots, which are 2 and 6.

Exam Tip: The word "hence" indicates that you must use your formulated polynomial to find the zeroes. Always follow this exact order of steps.

Question 8. If \( \sqrt{3} \) and \( -\sqrt{3} \) are two of the zeros of the polynomial \( f(x) = x^4-3x^3-x^2+9x-6 \), find the other zeros.
Answer: Since \( \sqrt{3} \) and \( -\sqrt{3} \) are zeroes, we have:
\( (x - \sqrt{3})(x + \sqrt{3}) = x^2 - 3 \) as a factor of \( f(x) \).

Dividing \( f(x) \) by \( x^2 - 3 \) using polynomial division:
\[ \begin{array}{rll}
x^2 - 3x + 2 & \text{(Quotient)} \\
x^2 - 3 \ \overline{\big) \ x^4 - 3x^3 - \phantom{0}x^2 + 9x - 6} \\
\underline{-\left(x^4 \phantom{- 3x^3} - 3x^2\right)} \phantom{+ 9x - 6} \\
-3x^3 + 2x^2 + 9x - 6 \\
\underline{-\left(-3x^3 \phantom{+ 2x^2} + 9x\right)} \phantom{- 6} \\
2x^2 \phantom{+ 9x} - 6 \\
\underline{-\left(2x^2 \phantom{+ 9x} - 6\right)} \\
0 & \text{(Remainder)}
\end{array} \]
The other factor is \( x^2 - 3x + 2 \). Finding its roots by factoring:
\( x^2 - 3x + 2 = 0 \)
\( \implies (x - 1)(x - 2) = 0 \)
\( \implies x = 1 \) or \( x = 2 \).
Therefore, the other zeroes of the polynomial are \( 1 \) and \( 2 \).
In simple words: The two known roots give us the quadratic divisor \( x^2 - 3 \). Dividing the main expression by this factor leaves us with \( x^2 - 3x + 2 \), which splits to find our other roots: 1 and 2.

Exam Tip: Be sure to keep track of missing powers of x when writing down the division steps. Leaving spaces helps align identical exponent terms properly.

Question 9. Obtain all the zeros of the polynomial \( f(x) = 2x^4+x^3-14x^2-19x-6 \), if two of its zeros are -2 and -1
Answer: Since \( -2 \) and \( -1 \) are zeroes of the polynomial, we can form the quadratic factor:
\( (x + 2)(x + 1) = x^2 + 3x + 2 \).

Dividing \( f(x) \) by \( x^2 + 3x + 2 \) using long division:
\[ \begin{array}{rll}
2x^2 - 5x - 3 & \text{(Quotient)} \\
x^2 + 3x + 2 \ \overline{\big) \ 2x^4 + \phantom{0}x^3 - 14x^2 - 19x - 6} \\
\underline{-\left(2x^4 + 6x^3 + \phantom{0}4x^2\right)} \phantom{- 19x - 6} \\
-5x^3 - 18x^2 - 19x - 6 \\
\underline{-\left(-5x^3 - 15x^2 - 10x\right)} \phantom{- 6} \\
-3x^2 - \phantom{0}9x - 6 \\
\underline{-\left(-3x^2 - \phantom{0}9x - 6\right)} \\
0 & \text{(Remainder)}
\end{array} \]
The other factor of the polynomial is the quotient, \( 2x^2 - 5x - 3 \). Finding its roots:
\( 2x^2 - 5x - 3 = 0 \)
\( \implies 2x^2 - 6x + x - 3 = 0 \)
\( \implies 2x(x - 3) + 1(x - 3) = 0 \)
\( \implies (2x + 1)(x - 3) = 0 \)
\( \implies x = -\frac{1}{2} \) or \( x = 3 \).
Therefore, all the zeroes are \( -2 \), \( -1 \), \( -\frac{1}{2} \), and \( 3 \).
In simple words: The given zeroes help us build a dividing quadratic factor, \( x^2 + 3x + 2 \). Dividing our polynomial by it leaves the quotient \( 2x^2 - 5x - 3 \), which has roots at \( -\frac{1}{2} \) and \( 3 \).

Exam Tip: Be sure to factorise the quadratic quotient completely and list "all the zeroes" as requested, including the given ones.

Question 10. Two zeros of the polynomial \( ax^3+3x^2-bx-6 \) are -1 and -2, find the values of a and b Also find the third zero.
Answer: Let \( p(x) = ax^3 + 3x^2 - bx - 6 \). Since \( -1 \) and \( -2 \) are zeroes:

1. Substituting \( x = -1 \):
\( p(-1) = 0 \)
\( \implies a(-1)^3 + 3(-1)^2 - b(-1) - 6 = 0 \)
\( \implies -a + 3 + b - 6 = 0 \)
\( \implies -a + b = 3 \) (Equation 1)

2. Substituting \( x = -2 \):
\( p(-2) = 0 \)
\( \implies a(-2)^3 + 3(-2)^2 - b(-2) - 6 = 0 \)
\( \implies -8a + 12 + 2b - 6 = 0 \)
\( \implies -8a + 2b = -6 \)
\( \implies -4a + b = -3 \) (Equation 2)

Subtracting Equation 2 from Equation 1:
\( (-a + b) - (-4a + b) = 3 - (-3) \)
\( \implies 3a = 6 \)
\( \implies a = 2 \)

Substituting \( a = 2 \) into Equation 1:
\( -2 + b = 3 \)
\( \implies b = 5 \).
So, \( a = 2 \) and \( b = 5 \).

The cubic polynomial is \( 2x^3 + 3x^2 - 5x - 6 \). Let the third zero be \( \gamma \).
Using the sum of zeroes relationship:
\( \alpha + \beta + \gamma = -\frac{\text{Coefficient of } x^2}{\text{Coefficient of } x^3} \)
\( \implies -1 + (-2) + \gamma = -\frac{3}{2} \)
\( \implies -3 + \gamma = -\frac{3}{2} \)
\( \implies \gamma = 3 - \frac{3}{2} = \frac{3}{2} \).
Therefore, \( a = 2 \), \( b = 5 \), and the third zero is \( \frac{3}{2} \).
In simple words: Putting \( -1 \) and \( -2 \) into the polynomial creates two equations. Solving them gives \( a = 2 \) and \( b = 5 \). Then, using the sum-of-roots formula helps us find that the third zero is \( \frac{3}{2} \).

Exam Tip: Using the sum of zeroes relation is the easiest and fastest way to find the third root once the cubic coefficients are determined.

Question 11. Find all the zeros of the polynomial \( 2x^4-2x^3-7x^2+3x+6 \), if two zeros are \( -\sqrt{\frac{3}{2}} \) and \( \sqrt{\frac{3}{2}} \)
Answer: Given that the two zeroes are \( \pm\sqrt{\frac{3}{2}} \), we can form the quadratic factor:
\( \left(x + \sqrt{\frac{3}{2}}\right)\left(x - \sqrt{\frac{3}{2}}\right) = x^2 - \frac{3}{2} \).
To simplify, we multiply by 2 and use \( 2x^2 - 3 \) as our dividing factor.

Dividing the given polynomial by \( 2x^2 - 3 \):
\[ \begin{array}{rll}
x^2 - x - 2 & \text{(Quotient)} \\
2x^2 - 3 \ \overline{\big) \ 2x^4 - 2x^3 - 7x^2 + 3x + 6} \\
\underline{-\left(2x^4 \phantom{- 2x^3} - 3x^2\right)} \phantom{+ 3x + 6} \\
-2x^3 - 4x^2 + 3x + 6 \\
\underline{-\left(-2x^3 \phantom{- 4x^2} + 3x\right)} \phantom{+ 6} \\
-4x^2 \phantom{+ 3x} + 6 \\
\underline{-\left(-4x^2 \phantom{+ 3x} + 6\right)} \\
0 & \text{(Remainder)}
\end{array} \]
The other factor is the quotient, \( x^2 - x - 2 \). Setting it equal to zero to find the roots:
\( x^2 - x - 2 = 0 \)
\( \implies (x - 2)(x + 1) = 0 \)
\( \implies x = 2 \) or \( x = -1 \).
Therefore, all the zeroes are \( \pm\sqrt{\frac{3}{2}} \), \( 2 \), and \( -1 \).
In simple words: We find the quadratic factor \( 2x^2 - 3 \) from the given roots and divide our main polynomial by it. Factoring the resulting quadratic quotient \( x^2 - x - 2 \) yields the final roots: \( 2 \) and \( -1 \).

Exam Tip: Eliminating fractions from your quadratic factor before doing the long division makes the calculations much easier.

Question 12. Verify that \( \frac{1}{2} \), 1, -2 are zeros of the polynomial \( 2x^3+x^2-5x+2 \). Also verify that the sums of the zeros = \( -\frac{b}{a} \) and the product of the zeros = \( -\frac{d}{a} \)
Answer: Let \( p(x) = 2x^3 + x^2 - 5x + 2 \).
First, let us verify each number by substituting it into the polynomial:
1. For \( x = \frac{1}{2} \):
\( p\left(\frac{1}{2}\right) = 2\left(\frac{1}{8}\right) + \frac{1}{4} - 5\left(\frac{1}{2}\right) + 2 = \frac{1}{4} + \frac{1}{4} - \frac{5}{2} + 2 = \frac{1}{2} - \frac{5}{2} + 2 = -2 + 2 = 0 \). (Verified)

2. For \( x = 1 \):
\( p(1) = 2(1)^3 + (1)^2 - 5(1) + 2 = 2 + 1 - 5 + 2 = 0 \). (Verified)

3. For \( x = -2 \):
\( p(-2) = 2(-2)^3 + (-2)^2 - 5(-2) + 2 = -16 + 4 + 10 + 2 = 0 \). (Verified)

Verification of Relationships:
Let \( \alpha = \frac{1}{2} \), \( \beta = 1 \), and \( \gamma = -2 \). Comparing with standard form, we have \( a = 2, b = 1, c = -5, d = 2 \).

1. Sum of zeroes:
\( \alpha + \beta + \gamma = \frac{1}{2} + 1 - 2 = -\frac{1}{2} \)
Also, \( -\frac{b}{a} = -\frac{1}{2} \). (Verified)

2. Product of zeroes:
\( \alpha\beta\gamma = \frac{1}{2} \cdot 1 \cdot (-2) = -1 \)
Also, \( -\frac{d}{a} = -\frac{2}{2} = -1 \). (Verified)
In simple words: First we substitute the values to prove they all make the polynomial equal to zero. Then, we calculate the sum and product of these zeroes and show that they match the coefficient ratio formulas.

Exam Tip: Show the steps for both substitutions and relationship verifications clearly, as teachers award marks for each part separately.

Question 13. \( x^4 + x^3 + 8x^2 + ax + b \) is exactly divisible by \( x^2+1 \). Find the values of a and b
Answer: Since the polynomial is exactly divisible by \( x^2 + 1 \text{ish} \), the remainder must be 0.
Let us perform polynomial long division:
\[ \begin{array}{rll}
x^2 + x + 7 & \text{(Quotient)} \\
x^2 + 1 \ \overline{\big) \ x^4 + x^3 + 8x^2 + ax + b} \\
\underline{-\left(x^4 \phantom{+ x^3} + x^2\right)} \phantom{+ ax + b} \\
x^3 + 7x^2 + ax + b \\
\underline{-\left(x^3 \phantom{+ 7x^2} + x\right)} \phantom{+ b} \\
7x^2 + (a - 1)x + b \\
\underline{-\left(7x^2 \phantom{+ (a - 1)x} + 7\right)} \\
(a - 1)x + (b - 7) & \text{(Remainder)}
\end{array} \]
Since the remainder must be zero:
1. \( a - 1 = 0 \implies a = 1 \)
2. \( b - 7 = 0 \implies b = 7 \)
Therefore, the values are \( a = 1 \) and \( b = 7 \).
In simple words: We divide the main polynomial by \( x^2 + 1 \). Setting the leftover remainder terms to zero gives us \( a = 1 \) and \( b = 7 \).

Exam Tip: Be mindful of missing powers of x in the divisor (no x-term in \( x^2+1 \)). Leave empty gaps during subtraction to avoid adding terms with different powers.

Question 14. If \( \alpha \) and \( \beta \) are the zeros of the polynomial \( x^2-8x+15 \), find the value of \( \frac{1}{\alpha} + \frac{1}{\beta} \) without finding the zeros
Answer: For the quadratic polynomial \( x^2 - 8x + 15 \):
\( a = 1, b = -8, c = 15 \)
Using the relationship between zeroes and coefficients:
Sum of zeroes, \( \alpha + \beta = -\frac{b}{a} = -\frac{-8}{1} = 8 \)
Product of zeroes, \( \alpha\beta = \frac{c}{a} = \frac{15}{1} = 15 \)

Now, rewrite the given expression:
\( \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} \)
Substituting our sum and product values:
\( = \frac{8}{15} \).
Thus, the value of the expression is \( \frac{8}{15} \).
In simple words: The combined fraction simplifies to the sum of roots over their product. Substituting our sum (8) and product (15) yields \( \frac{8}{15} \) directly.

Exam Tip: Read questions carefully. If the prompt says "without finding the zeroes," do not factorise the polynomial, or you may lose marks.

Question 15. Divide \( f(x) = 14x^3-5x^2+9x+1 \) by \( g(x) = 2x-1 \) and obtain the quotient q(x) and the rem. r(x) Verify the division algorithm \( f(x) = g(x) \cdot q(x) + r(x) \)
Answer: Let us divide \( 14x^3 - 5x^2 + 9x + 1 \) by \( 2x - 1 \) using polynomial division:
\[ \begin{array}{rll}
7x^2 + x + 5 & \text{(q(x))} \\
2x - 1 \ \overline{\big) \ 14x^3 - 5x^2 + 9x + 1} \\
\underline{-\left(14x^3 - 7x^2\right)} \phantom{+ 9x + 1} \\
2x^2 + 9x + 1 \\
\underline{-\left(2x^2 - \phantom{0}x\right)} \phantom{+ 1} \\
10x + 1 \\
\underline{-\left(10x - 5\right)} \\
6 & \text{(r(x))}
\end{array} \]
Thus, the quotient is \( q(x) = 7x^2 + x + 5 \) and the remainder is \( r(x) = 6 \).

Verification of the Division Algorithm:
We must verify that \( f(x) = g(x) \cdot q(x) + r(x) \).
\( \text{RHS} = (2x - 1)(7x^2 + x + 5) + 6 \)
\( = 2x(7x^2 + x + 5) - 1(7x^2 + x + 5) + 6 \)
\( = 14x^3 + 2x^2 + 10x - 7x^2 - x - 5 + 6 \)
\( = 14x^3 - 5x^2 + 9x + 1 = f(x) \).
Since LHS = RHS, the division algorithm is verified.
In simple words: Long division gives a quotient of \( 7x^2 + x + 5 \) and a remainder of 6. Multiplying the divisor by our quotient and adding 6 returns our original polynomial, verifying our division.

Exam Tip: Be sure to write out the full multiplication steps during verification to show the examiner you have carried out the check rigorously.

Question 16. If one zero of the polynomial \( x^2-4x+1 \) is \( 2+\sqrt{3} \), find the other
Answer: Let the zeroes of the quadratic polynomial be \( \alpha = 2 + \sqrt{3} \) and \( \beta \).
Using the relationship between coefficients and zeroes:
Sum of zeroes, \( \alpha + \beta = -\frac{b}{a} \)
\( \implies 2 + \sqrt{3} + \beta = -\frac{-4}{1} \)
\( \implies 2 + \sqrt{3} + \beta = 4 \)
\( \implies \beta = 4 - (2 + \sqrt{3}) \)
\( \implies \beta = 2 - \sqrt{3} \).
Thus, the other zero is \( 2 - \sqrt{3} \).
In simple words: Since the sum of the roots is 4 and one root is \( 2+\sqrt{3} \), we subtract this root from 4 to find the other root, which is \( 2-\sqrt{3} \).

Exam Tip: Since irrational roots of polynomials with rational coefficients always occur in conjugate pairs, if one root is \( a + \sqrt{b} \), the other must be \( a - \sqrt{b} \).