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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