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 - $\frac{3 - 3\sqrt{5}}{2}$)

● Factorization

• Polynomial
• Polynomial Equation and its Roots
  
• Division Algorithm
  
• Remainder Theorem
  
• Problems on Remainder Theorem
  
• Factors of a Polynomial
  
• Worksheet on Remainder Theorem
  
• Factor Theorem
  
• Application of Factor Theorem

10th Grade Math

From Application of Factor Theorem to HOME