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 - \(\frac{3 - 3\sqrt{5}}{2}\))
● Factorization
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