How can I express the equation x² + 6x + 7 as x(a² + b) where a and b are integers?

To rewrite the quadratic expression x² + 6x + 7 in the form x(a² + b) where a and b are integers, we first need to understand the components of the expression.

The standard form of a quadratic equation is Ax² + Bx + C, where A, B, and C are constants. In our case, we have:

  • A = 1 (the coefficient of x²)
  • B = 6 (the coefficient of x)
  • C = 7 (the constant term)

To match the required form x(a² + b), we can factor the expression but notice that x(a² + b) suggests we are looking for a common variable x out front. In fact, this doesn’t lend itself to a straightforward factorization in this case, as we usually represent quadratics directly as polynomials rather than through this form.

However, we can certainly express the above expression using a slightly different perspective. Let’s complete the square or reorganize it for a better look:

1. Start with the expression:

    x² + 6x + 7

2. Complete the square on the first two terms:

   &nbsp=x² + 6x = (x + 3)² – 9

3. Substitute back into the original expression:

    (x + 3)² – 9 + 7 = (x + 3)² – 2

This is not exactly in the form of x(a² + b), but we relate it as best as we can by realizing we want to express this characteristic of leading with x. Note that with -2 being a constant, our a and b may lead us elsewhere in integer representation.

If we want to play around, we could say:

    x((x + 3) + (-2/(x + 3))) = (x)(a + b), keeping integers would mean we cannot proceed with such transformations neatly as that ultimately leads to rationals.

Thus, we can validate that the expression x² + 6x + 7 might not perfectly fit a classic form of x(a² + b) but serves as a great representation showing quadratics nonetheless.

In conclusion, while restructuring into x(a² + b) is intriguing, the simplest way remains x² + 6x + 7 given conventional quadratic manipulation without overstepping rational boundaries superficially.

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