1. Find the value of the polynomial \(5x – 4x^2 + 3\) at

1. \(x = 0\)
   $$\begin{aligned}p\left( x\right) &=5x-4x^2+3\\ p\left( 0\right) &=5-4\cdot({0})^2+3\\ &=5\cdot 0\cdot - 4\cdot 0+3\\ &=0+0+3\\ &=3\end{aligned}$$
2. \(x = -1\)
   $$\begin{aligned}p\left( -1\right) &=5\times \left( -1\right)-4\times \left( {-1}^2\right) +3\\ &=-5\cdot 1-4\cdot 1+3\\ &=-5-4+3\\ &=-9+3\\ &=-6\end{aligned}$$
3. \(x = 2\)
   $$\begin{aligned}p\left( 2\right) &=5\cdot 2-4\cdot (2)^2+3\\ &=5\cdot 2 -4\cdot 4 + 3\\ &=10-16+3\\ &=13-16\\ &=-3 \end{aligned}$$

2. Find \(p(0),~ p(1) ~and~ p(2)\) for each of the following polynomials:

1. \(p(y) = y^2 – y + 1\)
   • $$\begin{aligned} p\left( 0\right) &=0^{2}-0+1\\ &=1\end{aligned}$$
   • $$\begin{aligned}p\left( 1\right) &=1^{2}-1+1\\ &=1-1+1\\ &=1\end{aligned}$$
   • $$\begin{aligned}p\left( 2\right) &=2^{2}-2+1\\ &=4-2+1\\ &=5-2\\ &=3\end{aligned}$$
2. \(p\left( t\right)=2+t+2t^{2}-t^{3}\)
   • $$\begin{aligned} p\left( 0\right) &=2+0+2\cdot 0^{2}+0^{3}\\ &=2+0+0+0\\ &=2\end{aligned}$$
   • $$\begin{aligned}P\left( 1\right) &=2+1+2\cdot 1^{2}-1^{3}\\ &=2+1+2\cdot 1-1\\ &=2+1+2-1\\ &=5-1\\ &=4\end{aligned}$$
   • $$\begin{aligned}p\left( 2\right) &=2+2+2\cdot 2^{2}-2^{3}\\ &=2+2+2\cdot 4-8\\ &=2+2+8-8\\ &=4\end{aligned}$$
3. \(p\left( x\right) =x^{3}\)
   • $$\begin{aligned} p\left( 0\right) &=0^{3}\\ &=0\end{aligned}$$
   • $$\begin{aligned}P\left( 1\right) &=1^{3}\\ &=1\end{aligned}$$
   • $$\begin{aligned}P\left( 2\right) &=2^{3}\\ &=8\end{aligned}$$
4. \(p\left( x\right) =\left( x-1\right) \times \left( x+1\right)\)
   • $$\begin{aligned}p\left( 0\right) &=\left( 0-1\right) \times \left( 0+1\right) \\ &=-1\times 1\\ &=-1\end{aligned}$$
   • $$\begin{aligned}p\left( 1\right) &=\left( 1-1\right) \times \left( 1+1\right) \\ &=0\times 2\\ &=0\end{aligned}$$
   • $$\begin{aligned}p\left( 2\right) &=\left( 2-1\right) \times \left( 2+1\right) \\ &=1\cdot 3\\ &=3\end{aligned}$$

3. Verify whether the following are zeroes of the polynomial, indicated against them.

1. $$\begin{aligned}p\left( x\right) &=3x+1,x=-\frac{1}{3}\\ P\left( \frac{-1}{3}\right) &=3\cdot \left( -\frac{1}{3}\right) +1\\ &=-3\times \dfrac{1}{3}+1\\ &=-1+1\\ &=0\end{aligned}$$ hence, \(x=-\frac{1}{3}\) is zero of polynomial.
2. $$\begin{aligned}p\left( x\right) &=5x-\pi ,x=4/5\\ P\left( 4/5\right) &=5\times \dfrac{4}{5}-\pi \\ &=4-\pi\\ &\neq 0\end{aligned}$$ hence, \(x=4/5\) is not a zero of the polynomial.
3. $$\begin{aligned}p\left( x\right) &=x^{2}-1,x=1,-1\\ p\left( 1\right) &=1^{2}-1\\ &=0\\\\ p\left( -1\right)&=\left( -1\right) ^{2}-1\\ &=1-1\\ &=0\end{aligned}$$ hence, \(x=1~and ~x=-1\) are zero of polynomial
4. $$\begin{aligned}p\left( x\right) &=\left( x+1\right) \times \left( x-2\right) ,x=-1,2\\ p\left( -1\right) &=\left( -1+1\right) \times \left( -1-2\right) \\ &=0\times \left( -3\right) \\ &=0\\\\ p\left( 2\right) &=\left( 2+1\right) \times \left( 2-2\right) \\ &=3\times 0\\ &=0\end{aligned}$$ hence, \(x=1~and ~x=2\) are zero of polynomial
5. $$\begin{aligned}p\left( x\right) &=x^{2},x=0\\ p\left( 0\right) &=0^{2}\\ &=0\end{aligned}$$ hence, \(x=0\) is zero of polynomial
6. $$\begin{aligned}p\left( x\right) =lx+m,\quad x&=-\dfrac{m}{l}\\ p\left( \frac{-m}{l }\right) &=l\left( \dfrac{-m}{l}\right) +m\\ &=-\dfrac{l\cdot m}{l}+m\\ &=-m+m\\ &=0\end{aligned}$$ hence, \(x=-\dfrac{m}{l}\) is zero of polynomial
7. $$\begin{aligned}p\left( x\right) &=3x^{2}-1,\quad x=\dfrac{-1}{\sqrt{3}},\dfrac{2}{\sqrt{3}}\\ p\left( -\dfrac{1}{\sqrt{3}}\right) &=3\cdot \left( -\dfrac{1}{\sqrt{3}}\right) ^{2}-1\\ &=3\cdot \dfrac{1}{3}-1\\ &=1-1\\ &=0\\\\ p\left( \dfrac{2}{\sqrt{3}}\right) &=3\cdot \left( \dfrac{2}{\sqrt{3}}\right) ^{2}-1\\ &=3\cdot \dfrac{4}{3}-1\\ &=4-1\\ &=3\end{aligned}$$ hence, \(x=\dfrac{-1}{\sqrt{3}}\) is zero of polynomial but \(x=\dfrac{2}{\sqrt{3}}\) is not a zero of polynomial
8. $$\begin{aligned}p\left( x\right) &=2x+1,\quad x=\frac{1}{2}\\ p\left(\frac{1}{2}\right) &=2\cdot \dfrac{1}{2}+1\\ &=1+1\\ &=2\end{aligned}$$ hence, \(x=\dfrac{1}{2}\) is not a zero of polynomial

4. Find the zero of the polynomial in each of the following cases:

1. \(p(x) = x + 5\)
   $$\begin{aligned}p\left( x\right) &=x+5\\ x+5&=0\\ x&=-5\end{aligned}$$
2. \( p(x) = x – 5\\\) $$\begin{aligned}p\left( x\right) &=x-5\\ x-5&=0\\ x&=5\end{aligned}$$
3. \(p(x) = 2x + 5\\\) $$\begin{aligned}p\left( x\right) &=2x+5\\ 2x+5&=0\\ 2x&=-5\\ x&=-\frac{5}{2}\end{aligned}$$
4. \( p(x) = 3x – 2\\\) $$\begin{aligned}p\left( x\right) &=3x-2\\ 3x-2&=0\\ 3x&=2\\ x&=\frac{2}{3}\end{aligned}$$
5. \( p(x) = 3x\\\) $$\begin{aligned}p\left( x\right) &=3x\\ 3x&=0\\ x&=0\end{aligned}$$
6. \( p(x) = ax,\quad a ≠ 0\\\) $$\begin{aligned}p\left( x\right) &=ax,\quad a\neq 0\\ ax&=0\\ x&=0\end{aligned}$$
7. \(p(x) = cx + d, \quad c ≠ 0, c, d are real numbers.\\\) $$\begin{aligned} cx+d&=0\\ cx&=-d\\ x&=-\dfrac{d}{c}\end{aligned}$$