Calculate QAP step by step for the given equation. Fill in the values in this markdown to complete the assignment

$$ x^2-x-42 == 0 $$

• $x * x = sym_1$
• $x + 42 = sym_2$
• $(sym_2)(-1) = sym_3$
• $sym_1 + sym_3 = out$

symbols = [~one, x, out, $sym_1$, $sym_2$, $sym_3$]

$\vec{s}$ = [1, 7, 0, 49, 49, -49]

$$ \vec{a} = [0, 1, 0, 0, 0, 0] $$

$$ \vec{b} = [0, 1, 0, 0, 0, 0] $$

$$ \vec{c} = [0, 0, 0, 1, 0, 0] $$

$$ \vec{a} = [42, 1, 0, 0, 0, 0] $$

$$ \vec{b} = [1, 0, 0, 0, 0, 0] $$

$$ \vec{c} = [0, 0, 0, 0, 1, 0] $$

$$ \vec{a} = [0, 0, 0, 0, 1, 0] $$

$$ \vec{b} = [-1, 0, 0, 0, 0, 0] $$

$$ \vec{c} = [0, 0, 0, 0, 0, 1] $$

$$ \vec{a} = [0, 0, 0, 1, 0, 1] $$

$$ \vec{b} = [1, 0, 0, 0, 0, 0] $$

$$ \vec{c} = [0, 0, 1, 0, 0, 0] $$

$$ A = \begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0 \\ 42 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 1 \\ \end{bmatrix} $$

$$ B = \begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ \end{bmatrix} $$

$$ C = \begin{bmatrix} 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ \end{bmatrix} $$

$$ A(x) = \begin{bmatrix} 21x^3 -168x^2 + 399x - 252 \\ \\ \frac{1}{3}x^3 - \frac{5}{2}x^2 + \frac{31}{6}x - 2 \\ \\ 0 \\ \\ \frac{1}{6}x^3 - x^2 + \frac{11}{6}x - 1 \\ \\ \frac{-1}{2}x^3 + \frac{7}{2}x^2 - 7x + 4 \\ \\ \frac{1}{6}x^3 - x^2 + \frac{11}{6}x - 1 \\ \\ \end{bmatrix} $$

$$ B(x) = \begin{bmatrix} \frac{7}{6}x^3 -\frac{17}{2}x^2 + \frac{55}{3}x - 11 \\ \\ \frac{-1}{6}x^3 + \frac{3}{2}x^2 - \frac{13}{3}x + 4 \\ \\ 0 \\ \\ 0 \\ \\ 0 \\ \\ 0 \\ \\ \end{bmatrix} $$

$$ C(x) = \begin{bmatrix} 0 \\ \\ 0 \\ \\ \frac{1}{6}x^3 - x^2 + \frac{11}{6}x - 1 \\ \\ \frac{-1}{6}x^3 + \frac{3}{2}x^2 - \frac{13}{3}x + 4 \\ \\ \frac{1}{2}x^3 -4x^2 + \frac{19}{2}x - 6 \\ \\ \frac{-1}{2}x^3 + \frac{7}{2}x^2 - 7x + 4 \\ \\ \end{bmatrix} $$

$$A(x).\vec{s} = \frac{-7}{6}x^3 -14x^2 + \frac{553}{6}x - 70$$

$$B(x).\vec{s} = 2x^2 -12x + 17$$

$$C(x).\vec{s} = \frac{245}{6}x^3 -294x^2 + \frac{3577}{6}x - 294$$

$$A(x).\vec{s} * B(x).\vec{s} - C(x).\vec{s} = \frac{7}{3}x^5 -14x^4 + \frac{875}{3}x^3 - 1190x^2 + \frac{5432}{3}x - 896$$ $$A(x).\vec{s} * B(x).\vec{s} - C(x).\vec{s} = \frac{-7}{3}(x-1)(x-2)(x-3)(x-4)(x+16)$$

Since the above polynomial is equal to $H(x).Z(x)$ it should have roots at x = 1, 2, 3, 4. Verify the same by pasting the polynomial here.

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  git clone https://github.com/DappCamp-Assignments/f79ca319bd4c68503d8ccb662126d4a5f82098a05519b1585b9c11b5eca39551
  

• Create a new branch with your name. You can use the following command

  git checkout -b satyajeet-sindhiyani
  

• Fill in the values in this file with your solution

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