(2)

Let $p\left(x\right)=x^2-3x-m(m+3)$

$\Rightarrow p\left(x\right)=x^2-(m+3)x+mx-m(m+3)$

$=x\{x-(m+3\left)\right\}+m\{x-(m+3\left)\right\}$

For zeros of $p\left(x\right)$

$\Rightarrow p\left(x\right)=(x+m)\{x-(m+3\left)\right\}=0$$\Rightarrow x=-m,m+3$

$\therefore$ Its zeros are $-m,m+3$

(2)

Given, $f\left(x\right)=p{x}^2-2x+3p$

Since $\alpha$ and $\beta$ are the zeroes of the given polynomial.

$\therefore$ $\alpha +\beta =-\frac{(-2)}{p}=\frac{2}{p}, \text{and} \alpha \beta =\frac{3p}{p}=3$

$\because$ $\alpha +\beta =\alpha \beta$      (given)

$\therefore$ $\frac{2}{p}=3\Rightarrow p=\frac{2}{3}$

(1)

Given polynomials : $x^2+px+q$ ...(i)

and $4x^2-5x-6$       ...(ii)

Zero of polynomial $4x^2-5x-6$ are: $x=2$ and $x=-3/4$

Now, zero of polynomial $x^2+px+q$ are 4 and $-3/2$

$\therefore$ Sum of zeroes $=-\frac{p}{1}$

$\Rightarrow 4-\frac{3}{2}=-\frac{p}{1}\Rightarrow -p=\frac{5}{2}\Rightarrow p=-\frac{5}{2}$

(2)

Sum of zeroes $=\sqrt{2}=\frac{-(-k\sqrt{2})}{2}\Rightarrow k=2$

(3)

$\alpha +\beta =\frac{-b}{a}=\frac{-3}{5}, \alpha \beta =\frac{c}{a}=\frac{-7}{5}$

Now, $\frac{1}{\alpha }+\frac{1}{\beta }=\frac{\alpha +\beta }{\alpha \beta }=\frac{\frac{-3}{5}}{\frac{-7}{5}}=\frac{3}{7}$

(3)

$$\begin{gathered} p\left(x\right) = x³ – 3x² + 2x \\ \Rightarrow p\left(x\right) = x(x² – 3x + 2) \\ \Rightarrow p\left(x\right) = x(x – 2)(x – 1) \end{gathered}$$

Factoring p(x) gives x(x – 1)(x – 2), indicating zeros at x = 0, 1, and 2. The graph intersects the x-axis at these points.

(1)

Given polynomial r(x) can be rewritten as (x – 2)², indicating a repeated zero at x = 2. The graph touches the x-axis at this point and does not cross it. The vertex of the graph is at point (2, 0).

(2)

At x = 1, t(1) = 1 – 6 + 11 – 6 = 0

So (x – 1) is a factor of t(x).

∴ t(x) = (x – 1)(x² – 5x + 6)

⇒ t(x) = (x – 1)(x – 2)(x – 3)

Factoring t(x) as (x – 1)(x – 2)(x – 3) reveals zeros at x = 1, 2, 3. The graph intersects the x-axis at these points.

(1)

Since the given graph intersects the x-axis at x = –2, therefore (x + 2) is a factor of h(x).

So, on dividing h(x) by (x + 2), we get remainder zero

(2)

Given polynomial is p(x) = 2x² – k√2x + 1.

Let α, β be its zeros then

α + β = –b/a = –(–k√2)/2 = k√2/2 = √2

[Here a = 2, b = –k√2, c = 1]

⇒ k√2 = 2√2 ⇒ k = 2

(4)

Given polynomial is p(x) = kx² – 30x + 45k.

We know α + β = –b/a and αβ = c/a

⇒ α + β = 30/k and αβ = 45k/k = 45

Given α + β = αβ ⇒ 30/k = 45 ⇒ k = 30/45 = 2/3

(1)

Given quadratic polynomial is 2x² – 3x – 9.

For zeros, we have 2x² – 3x – 9 = 0 ⇒ 2x² – 6x + 3x – 9 = 0 ⇒ (x – 3)(2x + 3) = 0

⇒ x = 3 or x = –3/2

(4)

Given: zeros of the polynomial are –5 and –3.

Polynomials having zeros –5 and –3 are given by p(x) = k(x + 5)(x + 3).

⇒ p(x) = k(x² + 8x + 15), where k is any real constant.

∴ k can have infinitely many values. So, number of possible polynomials is infinitely many.

(1)

Given: one zero = 2.

∴ Substituting x = 2, we get 2² + 3(2) + k = 0 ⇒ 4 + 6 + k = 0 ⇒ k = –10

(1)

The degree of polynomial having zeros –3 and 4, i.e., only two zeros, is 2 because it will be a quadratic polynomial.

i.e., p(x) = (x – (–3))(x – 4) ⇒ p(x) = (x + 3)(x – 4) ⇒ p(x) = x² – x – 12, which is a polynomial of degree 2.

(3)

Divide dividend by divisor, we get:

x – 1 | 2x³ + 3x² – 9x + 12
        2x² + 5x – 4

So, the quotient is 2x² + 5x – 4 and remainder is 8.

(2)

Since p is a prime number, its factors are 1 and p itself. So, α = 1 and β = p.

General form of quadratic equation: x² – (sum of zeros)x + product of zeros = 0.

⇒ x² – (1 + p)x + p = 0 is the required quadratic equation.

(2)

Given polynomial is f(x) = px² – 2x + 3p.

So, α + β = –b/a = –(–2)/p = 2/p and αβ = c/a = 3p/p = 3 [a = p, b = –2, c = 3p].

Since α + β = αβ ⇒ 2/p = 3 ⇒ p = 2/3