**Q.1: If one zero of the polynomial \(x^{2} + x – 4x + 1\) is \((2+ \sqrt{3})\), write the other zero.**

**Sol:**

Let the other zeros of \(x^{2} + x – 4x + 1\)

By using the relationship between the zeros of the quadratic polynomial

We have, Sum of zeros = \(\frac{- (Coefficient \; of \; x )}{Coefficient \; of \; x^{2}}\)

\(Therefore, 2 + \sqrt{3} + a = \frac{-(-4)}{1}\)

\( \Rightarrow a = 2 – \sqrt{3}\)

Hence, the other zeros of \(x^{2} + x – 4x + 1\)

**Q.2: Find the zeros of the polynomial \(x^{2} + x – p(p + 1)\).**

**Sol:**

f(x) = \(x^{2} + x – p(p + 1)\)

By adding and subtracting px, we have

f(x) = \(x^{2} + px + x – px – p(p+1)\)

\(= x^{2} + (p+1) x – px – p(p+1)\)

\(= x \left [ x + (p+1) \right ] Â – p \left [ x + (p+1) \right ]\)

= \(= \left [ x + (p+1) \right ] (x – p)\)

\(f(x) = 0\)

\(\Rightarrow \left [ x + (p+1) \right ] (x – p) = 0\)

\(\Rightarrow \left [ x + (p+1) \right ] = 0 \;\; or \;\;(x – p) = 0\)

\(\Rightarrow x = -(p+1) \;\; or \;\;x = p\)

So, the zeros of f(x) are -(p+1) and p.

**Q.3: Find the zeros of the polynomial \(x^{2} – 3 x – m(m + 3)\).**

Sol:

f(x) = \(x^{2} – 3 x – m(m + 3)\)

By adding and subtracting mx , we have

f(x) = \(x^{2} – mx – 3 x + mx – m(m + 3)\)

\(= x \left [ x – (m+3) \right ](x + m) + m \left [ x – (m + 3) \right ]\)

\(= \left [ x -(m+3) \right ] Â (x + m)\)

f (x) = 0

\(\Rightarrow \left [ x -(m+3) \right ] Â (x + m) = 0\)

\(\Rightarrow \left [ x -(m+3) \right ] = 0 \;\; or \;\; (x + m) = 0\)

\(\Rightarrow x = m + 3 \;\; or \;\; x = -m\)

So, the zeros of f(x) are -m and m+3.

**Q.4: If \(\alpha \;\; and \;\; \beta\) are the zeros of the polynomial such that \(\alpha + \beta = 6\) and \(\alpha \beta = 4\)**

**Then write the polynomial.**

**Sol: **

If \(\alpha \;\; and \;\; \beta\)

\(x^{2} – (\alpha + \beta )x + \alpha \beta\)

Substituting the values from the given conditions

\(x^{2} – 6x + 4\)

**Q.5: If one zero of the Quadratic polynomial \(kx^{2} + 3x + k \;\; is \;\; 2\), then find the value of k.**

**Sol:**

If x =2 is one zero of the quadratic polynomial \(kx^{2} + 3x + k\)

Therefore, Â it will satisfy the above polynomial.

Now, we have

\(k(2)^{2} + 3(2) + k = 0\)

\(\Rightarrow 4k + 6 + k = 0\)

\(\Rightarrow 5k + 6 = 0\)

\(\Rightarrow k = -\frac{6}{5}\)

**Q.6: If 3 is a zero of the polynomial \( 2x^{2} + x + k \;\; is \;\; 2\), then find the value of k .**

**Sol:**

Given x=3 is one zero of the polynomial \( 2x^{2} + x + k \)

Therefore it will satisfy the above polynomial.

Now, we have

\(2(3)^{2} + 3 + k = 0\)

\(\Rightarrow 21 + k = 0\)

\(\Rightarrow k = -21\)

**Q.7: If -4 is a zero of the polynomial \(x^{2} – x – (2k + 2)\) then find the value of k .**

**Sol:**

Given x= -4 is one zero of the polynomial \(x^{2} – x – (2k + 2)\)

Therefore it will satisfy the above polynomial.

Now, we have

\((-4)^{2} – (-4) – (2k + 2) = 0\)

\(\Rightarrow 16 + 4 – 2k – 2 = 0\)

\(\Rightarrow – 2k = -18\)

\(\Rightarrow k = 9\)

**Q.8: If 1 is a zero of the polynomial \(ax^{2} – 3(a-1)x -1\) then find the value of a.**

**Sol: **

Given x= 1 is one zero of the polynomial \(ax^{2} – 3(a-1)x -1\)

Therefore it will satisfy the above polynomial.

Now, we have

\(a(1)^{2} – 3(a-1)1 -1 = 0\)

\(\Rightarrow a – 3a + 3 – 1 = 0\)

\(\Rightarrow – 2a = -2\)

\(\Rightarrow a = 1\)

**Q.9: If -2 is a zero of the polynomial \( 3x^{2} + 4x + 2k\) then find the value of k.**

**Sol:**

Given x= -2 is one zero of the polynomial \(3x^{2} + 4x + 2k\)

Therefore it will satisfy the above polynomial.

Now, we have

\(3(-2)^{2} + 4 (-2) + 2k = 0\)

\(\Rightarrow 12 – 8 + 2k = 0 \)

\(\Rightarrow k = -2\)

**Q.10: Write the zeros of the polynomial \( x^{2} – x – 6\)**

**Sol:**

\(f(x) = x^{2} – x – 6\)

\(= x^{2} – 3x + 2x – 6\)

\(= x(x – 3) + 2( x- 3)\)

\(= (x – 3) ( x – 2)\)

\(f(x) = 0 \\ \Rightarrow (x – 3) ( x – 2) = 0 \\ \Rightarrow (x – 3) = 0 \;\; or \;\; (x-2) = 0 \\ \Rightarrow x = 3 \;\; or \;\; x = -2\)

**Q.11: If the sum of the zeros of the Quadratic polynomial \(kx^{2} – 3x + 5\) is 1, write the value of k.**

**Sol:**

By using the relationship between the zeros of the polynomial,

We have

Sum of zeros = \(\frac{- (Coefficient \; of \; x )}{Coefficient \; of \; x^{2}}\)

\(\Rightarrow 1 = \frac{- (-3)}{k}\)

\(\Rightarrow k = 3\)

**Q.12: If the product of the zeros of the Quadratic polynomial \( x^{2} – 4x + k\) is 3 then write the value of k.**

**Sol:**

By using the relationship between the zeros of the quadratic polynomial

We have

Product of zeros = \(\frac{Constant \; term}{ Coefficient \; of \; x^{2}}\)

\(\Rightarrow 3 = \frac{k}{1}\)

\(\Rightarrow k = 3\)

**Q.13: If (x+a) is a factor of \(2x^{2} + 2ax + 5x + 10\), find the value of Â a.**

**Sol:**

Given (x+a) is a factor of \(2x^{2} + 2ax + 5x + 10\)

We have

x + a = 0

x = -a

Since (x + a) is a factor of \(2x^{2} + 2ax + 5x + 10\)

Hence, it will satisfy the above polynomial

\(Therefore, Â 2(-a)^{2} + 2a (-a) + 5(-a) + 10 = 0\)

\(\Rightarrow -5a + 10 = 0\)

\(\Rightarrow a = 2\)

**Q.14: If (a – b) , a and ( a + b) area zeros of the polynomial \(2x^{3} – 6x^{2} + 5x – 7\) then write the value of a.**

**Sol:**

By using the relationship between the zeros of the cubic polynomial

We have,

Sum of zeros = \(\frac{- (Coefficient \; of \; x^{2} )}{Coefficient \; of \; x^{3}}\)

\(\Rightarrow a – b + a + a + b = \frac{-(-6)}{2}\)

\(\Rightarrow 3a = 3\)

\(\Rightarrow a = 1\)

**Q.15: If \(x^{3} + x^{2} – ax + b\) is divisible by \(x^{2} – x\), write the value of x.**

**Sol:**

Equating \(x^{2} – x\)

\(x (x – 1) = 0\)

\(\Rightarrow x = 0 \;\; or \;\; (x – 1) = 0\)

\(\Rightarrow x = 0 \;\; or \;\; x = 1\)

Since \(x^{3} + x^{2} – ax + b\)

Hence, the zeros of \(x^{2} – x\)

\(Therefore, Â (0)^{3} + 0 ^{2} – a(0) + b = 0\)

\(\Rightarrow b = 0\)

And

\((1)^{3} + 1 ^{2} – a(1) + 0 = 0 \;\;\;\; [Since, b = 0]\)

\(\Rightarrow a = 2\)

**Q.16: Â If \(\alpha \;\; and \;\; \beta\) are the zeros of the polynomial \(2x^{2} + 7x + 5\), write the value of \(\alpha + \beta + \alpha \beta\) .**

**Sol:**

By using the relationship between the zeros of the quadratic polynomial

We have,

Sum of zeros = \(\frac{- (Coefficient \; of \; x )}{Coefficient \; of \; x^{2}}\)

Product of zeros = \(\frac{- (Constant \; term )}{Coefficient \; of \; x^{2}}\)

\(Therefore, Â \alpha + \beta = \frac{-7}{2} \;\; and \;\; \alpha \beta = \frac{-7}{5} + \frac{5}{2}\)

Now, \(Therefore, Â \alpha + \beta + \alpha \beta = \frac{-7}{2} + \frac{5}{2} = -1\)

**Q.17: State division algorithm for the polynomial.**

**Sol:**

If f(x) and g(x) are two polynomial such that degree of f(x) is greater than degree of g(x) where g(x) \( \neq\)

\(f(x) = g(x) \times q(x) + r(x)\)

Where r(x) = 0 or degree of r(x) \(<\)

**Q.18: The sum of the zeros and the product of zeros of a Quadratic polynomial are \(\frac{-1}{2} \;\; and \;\; -3\) respectively. Write the polynomial.**

**Sol:**

We can find the quadratic polynomial if we know the sum of the roots and product of the roots by using the formula

\(x^{2}\)

\(\Rightarrow x^{2} – \left ( =\frac{1}{2} \right ) x + (-3)\)

\(\Rightarrow x^{2} + \frac{1}{2}x – 3\)

Hence the required polynomial is \(\Rightarrow x^{2} + \frac{1}{2}x – 3\)

**Q.19: Write the zeros of the Quadratic polynomial \(f(x) = 6x^{2} -3\)**

**Sol:**

To find the zeros of the quadratic polynomial we will equate f(x) = 0

\(Therefore, Â f(x) = 0\)

\(\Rightarrow 6x^{2} – 3 = 0\)

\(\Rightarrow 3 (2x^{2} – 1) = 0\)

\(\Rightarrow (2x^{2} – 1) = 0\)

\(\Rightarrow 2x^{2} = 1\)

\(\Rightarrow x^{2} = \frac{1}{2}\)

\(\Rightarrow x = \pm \frac{1}{\sqrt{2}}\)

Hence, the zero of the quadratic polynomial \(f(x) = 6x^{2} -3\)

**Q.20: Write the zeros of the Quadratic polynomial \(f(x) = 4\sqrt{3} x^{2} + 5x – 2\sqrt{3}\)**

**Sol:**

To find the zeros of the quadratic polynomial we will equate f(x) = 0

\(Therefore, Â f(x) = 0\)

\(\Rightarrow 4\sqrt{3} x^{2} + 5x – 2\sqrt{3} = 0\)

\(\Rightarrow 4\sqrt{3} x^{2} + 8x Â – 3x – 2\sqrt{3} = 0\)

\(\Rightarrow 4x (\sqrt{3} x + 2) – \sqrt{3} (\sqrt{3}x + 2) = 0\)

\(\Rightarrow (4x – \sqrt{3}) (\sqrt{3} x + 2) = 0 \)

\(\Rightarrow (4x – \sqrt{3}) = 0 \;\; or \;\; (\sqrt{3} x + 2) = 0\)

\(\Rightarrow x = \frac{-2}{\sqrt{3}} \;\; or \;\; x = \frac{\sqrt{3}}{4}\)

Hence, the zeros of the quadratic polynomial f(x) = \(f(x) = 4\sqrt{3} x^{2} + 5x – 2\sqrt{3}\)

**Q.21: Write the zeros of the Quadratic polynomial \(f(x) = x^{2} – 5x + k \) such that \(\alpha – \beta = 1\), find the value of k.**

**Sol:**

By using the relationship between the zeros of the quadratic equation polynomial:

We have,

Sum of zeros = Â \(\frac{- (Coefficient \; of \; x )}{Coefficient \; of \; x^{2}}\)

Product of zeros = \(\frac{- (Constant \; term )}{Coefficient \; of \; x^{2}}\)

\(Therefore, Â \alpha + \beta = \frac{-(-5)}{1} \;\; and \;\; \alpha \beta = \frac{k}{1}\)

\(\Rightarrow \alpha + \beta = 5 \;\; and \;\; \alpha \beta = \frac{k}{1}\)

Solving \(\alpha – \beta = 1 \;\; and \;\; \alpha + \beta = 5\)

\(\alpha = 3 \;\; and \;\; \beta = 2\)

Substituting the values of \(\alpha \beta = \frac{k}{1}\)

k = 6

**Q.22: If \(\alpha \;\; and \;\; \beta\) are zeros of the polynomial \(f(x) = 6x^{2} + x – 2 \), then find the value of \( \frac{\alpha}{\beta } + \frac{\beta}{\alpha }\).**

**Sol;**

By using the relationship between the zeros of the quadratic polynomials.

We have,

Sum of zeros = Â \(\frac{- (Coefficient \; of \; x )}{Coefficient \; of \; x^{2}}\)

Product of zeros = \(\frac{- (Constant \; term )}{Coefficient \; of \; x^{2}}\)

\( Therefore, Â \alpha + \beta = \frac{-(-1)}{6} \;\; and \;\; \alpha \beta = – \frac{1}{3}\)

\(\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha ^{2} + \beta^{2}}{\alpha \beta}\)

\(= \frac{ \alpha ^{2} + \beta^{2} + 2 \alpha \beta – 2 \alpha \beta }{\alpha \beta}\)

\(= \frac{ (\alpha ^{2} + \beta^{2})^{2} – 2 \alpha \beta }{\alpha \beta}\)

\(= \frac{ (- \frac{1}{6})^{2} – 2 \left ( -\frac{1}{6} \right )}{-\frac{1}{3}}\)

\(= \frac{ \frac{1}{36} + \frac{2}{3} } { \frac{2}{3}}\)

\(= – \frac{25} {12}\)

**Q.23: If \(\alpha \;\; and \;\; \beta\) are zeros of the polynomial \(f(x) = 5x^{2} – 7x + 1\), then find the value of \(\frac{1}{\alpha } + \frac{1}{\beta }\).**

**Sol:**

By using the relationship between the zeros of the quadratic polynomial.

We have,

Sum of zeros =Â \(\frac{- (Coefficient \; of \; x )}{Coefficient \; of \; x^{2}}\)

Product of zeros = \(\frac{- (Constant \; term )}{Coefficient \; of \; x^{2}}\)

\(Therefore, \alpha + \beta = \frac{-(-7)}{5} \;\; and \;\; \alpha \beta = – \frac{1}{5}\)

\(\Rightarrow \alpha + \beta = \frac{7}{5} \;\; and \;\; \alpha \beta = \frac{1}{5}\)

Now, \(\frac{1}{\alpha } + \frac{1}{\beta } = \frac{\alpha +\beta }{\alpha \beta }\)

\(= \frac{\frac{7}{5}}{\frac{1}{5}}\)

**Q.24: If \(\alpha \;\; and \;\; \beta\) are zeros of the polynomial \(f(x) = x^{2} + x – 2 \), then find the value of \(\frac{1}{\alpha } – \frac{1}{\beta }\).**

**Sol:**

By using the relationship between the zeros of the quadratic polynomial

We have,

Sum of zeros = Â \(\frac{- (Coefficient \; of \; x )}{Coefficient \; of \; x^{2}}\)

Product of zeros = \(\frac{- (Constant \; term )}{Coefficient \; of \; x^{2}}\)

\(\Rightarrow \alpha + \beta = \frac{-1}{1} \;\; and \;\; \alpha \beta = \frac{-2}{1}\)

\(\Rightarrow \alpha + \beta = -1 Â \;\; and \;\; \alpha \beta = -2 \)

Now, \(\left ( \frac{1}{\alpha } – \frac{1}{\beta } \right )^{2} = \left ( \frac{\beta – \alpha }{\alpha \beta } \right )^{2}\)

\(= \frac{\left ( \alpha + \beta \right )^{2} – 4 \alpha \beta }{\left ( \alpha \beta \right )^{2}} \;\;\;\;\; \left [Since, \left ( \beta -\alpha \right )^{2} = \left ( \alpha + \beta \right )^{2} – 4\alpha \beta \right ]\)

\(= \frac{ ( -1 )^{2} – 4 (-2) }{(-2)^{2}} \;\;\;\;\; \left [ Since, Â \alpha + \beta = -1 \;\; and \;\; \alpha \beta = -2 \right ]\)

\(= \frac{1 + 8}{4}\)

\(= \frac{9}{4}\)

\(Since, \left ( \frac{1}{\alpha } – \frac{1}{\beta } \right )^{2} = \frac{9}{4}\)

\(\Rightarrow \frac{1}{\alpha } – \frac{1}{\beta } = \pm \frac{3}{2}\)

**Q.25: If the zeros of the polynomial \(f(x) = x^{3} – 3x^{2} + x + 1\) are (a – b) , a and (a + b), find the value of a.**

**Sol:**

By using the relationship between the zeros of the cubic polynomial:

We have,

Sum of zeros = \(\frac{- (Coefficient \; of \; x )^{2}}{Coefficient \; of \; x^{3}}\)

Therefore, a – b + a + a + b = \(\frac{-(-3)}{1}\)

\(\Rightarrow 3a = 3\)

\(\Rightarrow a = 1\)

Now, Product of zeros = \(\frac{- (Constant \; term )}{Coefficient \; of \; x^{2}}\)

\(Therefore, (a – b) (a) (a + b) = \frac{-1}{1}\)

\(\Rightarrow (1 – b) (1) (1 + b) = -1 \left [ Since, a = 1 \right ]\)

\(\Rightarrow 1 – b^{2} = -1\)

\(\Rightarrow b^{2} = 2\)

\(\Rightarrow b = \pm \sqrt{2}\)