x2 + 5x + 4
(x+4)(x+1)
Why do we use the quadratic formula?
To find the roots/zeros of a parabola/graph.
(x2 + 2x - 1) + (x -3)
x2 + 3x - 4
(x2 + 2)(x+3)
x3 + 3x2 + 2x + 6
What is the first step of subtracting polynomials?
Distribute the negative sign.
n2 + 7n + 12
(n+4)(n+3)
What is another way to find the roots of a quadratic equation?
Factoring
(x2 - x) + (8x - 2x2)
-x2 + 7x
(x+4)(x-5)
x2 - x - 20
(5 + 5n3) - (1 - 3n3)
8n3+4
k2 - 13k + 40
(k-5)(k-8)
DOUBLE JEOPARDY
What is the quadratic formula?
x= -b +/- sq.rt(b2 - 4ac)/2a
(8n2 - 2n3) + (6n3 - 8n2)
4n3
(9x + 7)(x-4)
9x2 - 29x - 28
(2a2 + 4a3) - (3a3 + 8)
a3+ 2a2 - 8
2k2 + 22k +60
2(k+5)(k+6)
Use the quadratic formula to find the roots of the following equation: x2+3x-10=0
(2, 0) and (-5, 0)
(14 + 12a3) + (17a4 + 15 - 5a3)
17a4 + 7a3 + 29
(x2 + 2x3)(x +8)
2x4 + 17x3 + 8x2
(5x2+4) - (5 + 5x3)
-5x3 + 5x2 -1
5n2 + 10n +20
5(n2 + 2n + 4)
Use the quadratic formula to find the roots of the following equation: -x2 +9x + 10 =0
(-1, 0) and (10, 0)
(17v2- 8) + (17v2 + 10 + v3)
v3 + 34v2 + 2
(x4+2x)(7-x2)
-x6 + 7x4 - 2x3 + 14x
(10p4 + 11) - (11p4 + 13 + 16p2)
-p4 - 16p2 -2