Let P(x) represent a polynomial. To find the zeros of the polynomial, we have to set P(x) equal to zero and solve for x.
Find the zeros of each of the following polynomials.
Example 1 :
p(x) = 2x + 3
Solution :
p(x) = 0
2x + 3 = 0
Subtract 3 from both sides.
2x = -3
Divide both sides by 2.
x = -³⁄₂
Example 2 :
p(x) = 4x - 1
Solution :
p(x) = 0
4x - 1 = 0
Add 1 to both sides.
4x = 1
Divide both sides by 4.
x = -¼
Example 3 :
p(x) = x^{2} + 17x + 60
Solution :
p(x) = 0
x^{2} + 17x + 60 = 0
The above quadratic equation can be solved by factoring.
In the quadratic equation, 'x^{2} + 17x + 60 = 0', the coefficient of x^{2 }is 1 and the constant term is '+60'. We have to get two factors for '+60' such that the product of the two factors is '+60' and the sum of the two factors is '+17', which is the coeffient of x.
Such two factors of '+60' are '+12' and '+5'.
x^{2} + 17x + 60 = 0
x^{2} + 12x + 5x + 60 = 0
x(x + 12) + 5(x + 12) = 0
(x + 12)(x + 5) = 0
x + 12 = 0 or x + 5 = 0
x = -12 or x = -5
The zeros of the given quadratic polynomial are -12 and -5.
Example 4 :
p(x) = x^{2} - 4x + 1
Solution :
p(x) = 0
x^{2} - 4x + 1 = 0
In the quadratic equation, 'x^{2} - 4x + 1 = 0', the coefficient of x^{2 }is 1 and the constant term is '+1'. We have to get two factors for '+1' such that the product of the two factors is '+1' and the sum of the two factors is '-4', which is the coeffient of x.
But, we can not get such two factors for '+1'.
So, the quadratic equation 'x^{2} - 4x + 1 = 0' can not be solved by factoring.
We can solve this kind of quadratic equations using quadratic formula.
Comparing 'ax^{2} + bx + c = 0' and 'x^{2} - 4x + 1 = 0', we get
a = 1, b = -4 and c = 1
Quadratic Formula :
Substitute a = 1, b = -4 and c = 1.
x = 2 ± √3
x = 2 - √3 or 2 + √3
The zeros of the given quadratic polynomial are 2 - √3 and 2 + √3.
Example 5 :
p(x) = x^{3} - 5x^{2} - 4x + 20
Solution :
p(x) = x^{3} - 5x^{2} - 4x + 20
x^{3} - 5x^{2} - 4x + 20 = 0
The above cubic equation can be solved by factoring. The expression on the left side of the equation can be factored by grouping.
(x^{3} - 5x^{2}) + (-4x + 20) = 0
x^{2}(x - 5) - 4(x - 5) = 0
(x^{2} - 4)(x - 5) = 0
x^{2} - 4 = 0 x^{2} - 2^{2} = 0 (x + 2)(x - 2) = 0 x + 2 = 0 or x - 2 = 0 x = -2 or x = 2 |
x - 5 = 0 x = 5 |
The zeros of the given cubic polynomial are -2, 2 and 5.
Example 6 :
p(x) = 4x^{3} - 7x + 3
Solution :
p(x) = 4x^{3} - 7x + 3
4x^{3} - 7x + 3 = 0
The above cubic equation can not be solved by factoring. So, we can use synthetic division to solve the above cubic equation.
By trial and error, we can check 1 or -1 or 2 or -2..... can be the zeros of the given cubic polynomial using synthetic division.
When we divide the given cubic polynomial by (x - 1), the remainder is zero.
So, x = 1 is one of the zeros of the given cubic polynomial.
The quotient is (4x^{2} + 4x - 3).
Set the quotient equal to zero and solve for x.
4x^{2} + 4x - 3 = 0
Multiply the coefficent of x^{2}, 4 by the contant term -3. The product is -12. Now, we have to get two fators for -12 such that product is equal to -12 and the sum is equal to +4.
Such two factors are -2 and +6.
4x^{2} - 2x + 6x - 3 = 0
2x(2x - 1) + 3(2x - 1) = 0
(2x - 1)(2x + 3) = 0
2x - 1 = 0 or 2x + 3 = 0
x = ½ or x = -³⁄₂
The zeros of the given cubic polynomial are -³⁄₂, ½ and 1.
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