Quadratic Equation a)
b)
c)
d
e)
Its Factor Form
(x-3)(x - 1)=0
(x – 4)(x+2)=0
x(x+2)=0
(x-4)(x+4) = 0
(5x-3)(x+1) = 0
Its Solutions
x= 3 and x = 1
x= 4 and x = -2
x= 0 and x = -2
x= 4 and x = -4
x= -1 and x = 3/5

Explanation:

This is a product of two expressions that is equal to zero. Note that any xxx value that makes either (x-1)(x−1)left parenthesis, x, minus, 1, right parenthesis or (x+3)(x+3)left parenthesis, x, plus, 3, right parenthesis zero, will make their product zero.

begin{aligned} (x-1)&(x+3)=0

swarrow quad&quadsearrow x-1=0quad&quad x+3=0 x=1quad&quad x=-3 end{aligned}

(x−1)

↙

x−1=0

x=1

(x+3)=0

↘

x+3=0

x=−3

Substituting either x=1x=1x, equals, 1 or x=-3x=−3x, equals, minus, 3 into the equation will result in the true statement 0=0, equals, 0, so they are both solutions to the equation.

Bwould be my guess |-/ 

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