# Application of Factor Theorem

We will discuss here about the application of Factor Theorem.

1. Find the roots of the equation 2x$$^{2}$$ - 7x + 6 = 0. Hence factorize 2x$$^{2}$$ - 7x + 6.

Solution:

Here, the equation is 2x$$^{2}$$ - 7x + 6 = 0

⟹ 2x$$^{2}$$ - 4x - 3x + 6 = 0

⟹ 2x(x - 2) - 3(x - 2) = 0

⟹ (x - 2)(2x - 3) = 0

⟹ x - 2 = 0 or 2x - 3 = 0

⟹ x = 2 or x = $$\frac{3}{2}$$

Therefore, 2x$$^{2}$$ - 7x + 6 = 2(x - 2)(x - $$\frac{3}{2}$$) = (x - 2)(2x - 3)

2. Find the quadratic equation whose roots are 1 + √3 and 1 - √3.

Solution:

We know that the quadratic equation whose roots are α and β, is

(x – α)(x – β) = 0

Therefore, the required equation is {x - (1 + √3)}{x - (1 - √3)} = 0

⟹ x$$^{2}$$ - {1 - √3 + 1 + √3}x + (1 + √3)( 1 - √3) = 0

⟹ x$$^{2}$$ - 2x + (1 - 3) = 0

⟹ x$$^{2}$$ - 2x – 2 = 0.

3. Find the cubic equation whose roots are 2, √3 and -√3.

Solution:

We know that the quadratic equation whose roots are α, β and γ, is

(x – α)(x – β)(x - γ) = 0

Therefore, the required equation is (x – 2)(x - √3){x – (-√3)} = 0

⟹ (x - 2)(x - √3)(x + √3) = 0

⟹ (x - 2)(x$$^{2}$$ - 3) = 0

⟹ x$$^{3}$$ – 2x$$^{2}$$ - 3x + 6 = 0.

⟹ x$$^{2}$$ - {1 - √3 + 1 + √3}x + (1 + √3)( 1 - √3) = 0

⟹ x$$^{2}$$ - 2x + (1 - 3) = 0

⟹ x$$^{2}$$ - 2x - 2 = 0.

4. Factorize x$$^{2}$$ -3x - 9

Solution:

The corresponding equation is x$$^{2}$$ - 3x - 9= 0

Now we apply the quadratic formula

x = $$\frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$$

= $$\frac{-(-3) \pm \sqrt{(-3)^{2} - 4 \cdot 1 \cdot (-9)}}{2 \cdot 1}$$

= $$\frac{3 \pm \sqrt{9 + 36}}{2}$$

= $$\frac{3 \pm \sqrt{45}}{2}$$

= $$\frac{3 \pm 3\sqrt{5}}{2}$$

Therefore, x$$^{2}$$ - 3x - 9 = (x - $$\frac{3 + 3\sqrt{5}}{2}$$)(x - $$\frac{3 - 3\sqrt{5}}{2}$$)

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● Factorization

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