Thread: Fixed Real Numbers...Part 3

Suppose a, b, and c are fixed real numbers such that
b^2 - 4ac ≥ 0. Let r and s be the solutions of
ax^2 + bx + x = 0.

(b) Use part (a) to verify that ax^2 + bx + c =
a(x - r)(x - s)

Use part (b) to factor x^2 - 2x - 1 and 5x^2 + 8x + 2.

Originally Posted by sologuitar

Suppose a, b, and c are fixed real numbers such that
b^2 - 4ac ≥ 0. Let r and s be the solutions of
ax^2 + bx + x = 0.

(b) Use part (a) to verify that ax^2 + bx + c =
a(x - r)(x - s)

Use part (b) to factor x^2 - 2x - 1 and 5x^2 + 8x + 2.

??? What was "part a"? The quadratic formula?
Perhaps:
b) The fact that r and s satisfy means that (x- r) and (x- s) are factors. Since just multiplying them together gives , to get the "a" multiplying , we must multiply that product by a: a(x-r)(x-s)= ax^2- a(r+s)x+ ars= ax^2+ bx+ c.

c) Use the quadratic formula (or completing the square) to find the roots of each equation. Then use the fact, from (b) above, that the polynomial can be factored as a(x-r)(x-s) where a is the leading coefficient (1 in the first example, 5 in the second) and r and s are the solutions to the equations with the polynomial set to 0.

Originally Posted by HallsofIvy

??? What was "part a"? The quadratic formula?
Perhaps:
b) The fact that r and s satisfy means that (x- r) and (x- s) are factors. Since just multiplying them together gives [tex]x^2- (r+s)x+ rs, to get the "a" multiplying , we must multiply that product by a: a(x-r)(x-s)= ax^2- a(r+s)x+ ars= ax^2+ bx+ c.

c) Use the quadratic formula (or completing the square) to find the roots of each equation. Then use the fact, from (b) above, that the polynomial can be factored as a(x-r)(x-s) where a is the leading coefficient (1 in the first example, 5 in the second) and r and s are the solutions to the equations with the polynomial set to 0.

Yes, part a was the quadratic formula. Thanks for the extra tips.
I'll work on this a little later today.