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Find the HCF of x4+x3+x2+x+1 and x4+x2+1. Solve: 2x=y and 2y+5=3x.

Mathematics

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  • a) Find the HCF of x4+x3+x2+x+1x^4+x^3+x^2+x+1 and x4+x2+1x^4+x^2+1. Let P(x)=x4+x3+x2+x+1P(x) = x^4+x^3+x^2+x+1 and Q(x)=x4+x2+1Q(x) = x^4+x^2+1.

  1. Use the Euclidean algorithm to find the HCF.

    Divide P(x)P(x) by Q(x)Q(x). 1x4+x2+1)x4+x3+x2+x+1‾−(x4+x2+1)‾x3+x \begin{array}{r} 1 \\ x^4+x^2+1 \overline{) x^4+x^3+x^2+x+1} \\ \underline{-(x^4 \quad +x^2 \quad +1)} \\ x^3 \quad +x \\ \end{array} The remainder is R1(x)=x3+xR_1(x) = x^3+x.

  2. Divide Q(x)Q(x) by the remainder R1(x)R_1(x).

    $$
\begin{array}{r}
x \\
x^3+x \overline{) x^4 \quad +x^2 \quad +1} \\
\underline{-(x^4 \quad +x^2)} \\
1 \\
\end{array}
$$
The remainder is $R_2(x) = 1$.

  3. Since the remainder is a non-zero constant, the HCF is 1.

    The HCF of $x^4+x^3+x^2+x+1$ and $x^4+x^2+1$ is $\boxed{1}$.
    • b) Solve: 2x=y2x=y and 2y+5=3x2y+5=3x. We have a system of two linear equations:
      1. y=2xy = 2x
      2. 2y+5=3x2y + 5 = 3x

  4. Substitute the expression for yy from equation (1) into equation (2).

    $$
2(2x) + 5 = 3x
$$

  5. Simplify and solve for xx.

    $$
4x + 5 = 3x
$$
$$
4x - 3x = -5
$$
$$
x = -5
$$

  6. Substitute the value of xx back into equation (1) to find yy.

    $$
y = 2(-5)
$$
$$
y = -10
$$
The solution to the system of equations is $\boxed{x=-5, y=-10}$.
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