Roots that Come Out to 0

Jason76

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\(\displaystyle x(x^2 + 5x - 6) = 0\)

\(\displaystyle x(x - 1)(x + 6) = 0\)

Setting each linear factor to 0 and solving for x:

\(\displaystyle (x - 1) = 0, x = 1\)

\(\displaystyle (x + 6) = 0, x = -6 \)

\(\displaystyle x = 0, x = 1, x = -6\)

I see that the 1 and -6 came from the linear factors of

\(\displaystyle (x - 1)\) and \(\displaystyle (x+6)\)

But where did the "0 root" come from?
 
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In the case of

\(\displaystyle 3(x + 2)(x - 5) = 0\)

Would 3 be a linear factor and root?

No, it is neither.

Linear factors are 1st-degree polynomials, and 3 is not a polynomial.

Polynomial roots are values for the variable that cause the polynomial to evaluate to zero.

When x = 3, your example polynomial has a value of -30, not zero, so 3 is not a root.




It is better to think of 3 as the value of A in the standard form Ax^2 + Bx + C

In other words, it's the "leading coefficient" in this quadratic polynomial.

It's the same A that we see in vertex form A(x - h)^2 + k

Graphically, you may think of A as a "stretcher" or "condenser" because values other than 1 and -1 cause the parabola to horizontally widen or shrink.

The sign of A's value tells us whether the parabola opens upward or downward. (You will learn more about A when you study graph/function "transformations".)

Your example quadratic polynomial with A = 3 is 3x^2 - 9x - 30. It's graph is green.

The polynomials with A = 1 and A = -1 (x^2 - 3x - 10 and -x^2 + 3x + 10) are graphed in red and yellow, respectively.

When the leading coefficient A changes value from 1 to 3, the parabola shrinks horizontally.

3x^2-9x-30.JPG
 
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