# use synthetic division to show that the given x value is a zero of the polynomial. Then find all other zeros.?

P(x)=x^ 3 -2x^ 2 -5x+6; x = 1

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• You can show that 1 is a zero of the polynomial just by plugging it in for x.

P(1) = 1^3 - 2(1^2) - 5(1) + 6 = 1 - 2 - 5 + 6 = 0

If 1 is a zero, then (x-1) is a factor.

We can divide polynomials just like long division of numbers, with powers of x instead of decimal places. Negative coefficients are allowed - you never have to carry/borrow.

x^2 - x - 6

_______________

x-1 ) x^3 - 2x^2 - 5x + 6

x^3 - x^2

_______________

-x^2 - 5x + 6

-x^2 + x

____________

- 6x + 6

- 6x + 6

________

0

Synthetic division does exactly the same math as long division, but doesn't bother to write out all the x's.

-1 | 1 -2 -5 6  <- divisor and dividend

|   -1  1 6  <- coefficient of the last term in each line of the long division

____________

1 -1 -6 0  <- quotient and remainder

In any case, now you have P(x) / (x-1) = (x^2 - x - 6). If P(x) = 0 and x ≠ 1, then x^2 - x - 6 = 0.

You can guess at some other integers that might make this equation true, or just apply the quadratic formula.

• this picture should help • Setting up the synthetic division by putting a 1 on one side and all of the coefficients on the other:

1 | 1 -2 -5 6

-------------------

Drop the 1 down, multiply it by -1 and put it below the -2, then add the two and put that value below the -2:

1 | 1 -2 -5 6. . . . .1-------------------. . . 1 -1

Repeat the process until you are out of columns.  The end result should be zero. This shows that 1 is a root and (x - 1) is a factor:

1 | 1 -2 -5 6. . . . .1 -1 -6-------------------. . . 1 -1 -6 0

With the last value (the remainder) zero, the other values become coefficients of the quotient.  So your function factors to:

P(x) = x³ - 3x² - 5x + 6

P(x) = (x - 1)(x² - x - 6)

We can now factor the quadratic:

P(x) = (x - 1)(x - 3)(x + 2)

Now we know that all three roots are:

x = -2, 1, and 3