Decompose the polynomial into linear factors (x^2−7x+6)(x^2+3x−18) .(1 point) (x^2−7x+6)(x^2+3x−18)=

Answer 1

To decompose the polynomial \((x^2−7x+6)(x^2+3x−18)\) into linear factors, we first need to factor each quadratic polynomial separately.

1. Factoring \(x^2 - 7x + 6\): This can be factored by finding two numbers that multiply to \(6\) (the constant term) and add to \(-7\) (the coefficient of \(x\)). The numbers are \(-1\) and \(-6\): \[ x^2 - 7x + 6 = (x - 1)(x - 6) \]

2. Factoring \(x^2 + 3x - 18\): We need to find two numbers that multiply to \(-18\) and add to \(3\). The numbers are \(6\) and \(-3\): \[ x^2 + 3x - 18 = (x + 6)(x - 3) \]

Now we can substitute the factored forms back into the expression: \[ (x^2 - 7x + 6)(x^2 + 3x - 18) = (x - 1)(x - 6)(x + 6)(x - 3) \]

Thus, the complete factorization of the given polynomial into linear factors is: \[ (x - 1)(x - 6)(x + 6)(x - 3) \]

Answer 2

Can it be simplified anymore?

Answer 3

The expression \((x - 1)(x - 6)(x + 6)(x - 3)\) is already in its simplest form as a product of linear factors.

Each factor \((x - 1)\), \((x - 6)\), \((x + 6)\), and \((x - 3)\) is already linear, meaning it cannot be factored any further in the field of real numbers.

If you want to combine or rearrange terms, you can multiply the factors back together to obtain the original polynomial:

\[ ((x - 1)(x - 6))((x + 6)(x - 3)) \]

However, this process will only return you to the original quadratic form, which is not a simplification but a recomposition.

The factorization \((x - 1)(x - 6)(x + 6)(x - 3)\) is fully simplified as far as factoring into linear terms is concerned.