We will discuss about the factor theory of quadratic equation.

Suppose when we assume that β be a root of the quadratic equation ax\(^{2}\) + bx + c = 0, then we get (x - β) is a factor of the quadratic expression ax^2 + bx + c.

Conversely, when we assume that (x - β) is a factor of the quadratic expression ax\(^{2}\) + bx + c then β is a root of the quadratic equation ax\(^{2}\) + bx + c = 0.

**Proof:**

The given quadratic equation ax\(^{2}\) + bx + c = 0

According to the problem, β is a root of the quadratic equation

ax\(^{2}\) + bx + c = 0

Hence, aβ\(^{2}\) + bβ + c = 0

Now, ax\(^{2}\) + bx + c

= ax\(^{2}\) + bx + c - (aβ\(^{2}\) + bβ + c), [Since, aβ\(^{2}\) + bβ + c = 0]

= a(x\(^{2}\) - β\(^{2}\)) + b(x - β)

= (x - β)[a(x + β) + b]

Therefore, we clearly see that (x - β) is a factor of the quadratic expression ax\(^{2}\) + bx + c.

Conversely, when (x - α) is a factor of the quadratic expression ax\(^{2}\) + bx + c then we can express,

ax\(^{2}\) + bx + c = (x - α)(mx + n), where m (≠ 0) and n are constants.

Now, we need to substitute x = β on both sides of the identity ax\(^{2}\) + bx + c = (x - β)(mx + n) then we get,

aβ\(^{2}\) + bβ + c = (β - β) × (mβ + n) = 0

aβ\(^{2}\) + bβ + c = 0 × (mβ + n) = 0

It is evident that the equation ax\(^{2}\) + bx + c = 0 is satisfied by x = β.

Therefore, β is a root of the equation ax\(^{2}\) + bx + c = 0.

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