Problem 1: Zeros: -1, 1, 2; Degree: 3<\/h3>\n\n\n\n
To form a polynomial from the given zeros, we use the fact that if a number ( r ) is a zero of a polynomial, then ( (x – r) ) is a factor of the polynomial.<\/p>\n\n\n\n
Given zeros: ( -1, 1, 2 ), the corresponding factors are:<\/p>\n\n\n\n
- \n
- ( (x + 1) ) for zero -1<\/li>\n\n\n\n
- ( (x – 1) ) for zero 1<\/li>\n\n\n\n
- ( (x – 2) ) for zero 2<\/li>\n<\/ul>\n\n\n\n
Thus, the polynomial is:<\/p>\n\n\n\n
[
f(x) = (x + 1)(x – 1)(x – 2)
]<\/p>\n\n\n\nNext, expand this expression step by step.<\/p>\n\n\n\n
- \n
- First, expand ( (x + 1)(x – 1) ) using the difference of squares formula:
[
(x + 1)(x – 1) = x^2 – 1
]<\/li>\n\n\n\n - Now multiply ( (x^2 – 1) ) by ( (x – 2) ):
[
(x^2 – 1)(x – 2) = x^3 – 2x^2 – x + 2
]<\/li>\n<\/ol>\n\n\n\nSo the final polynomial is:
[
f(x) = x^3 – 2x^2 – x + 2
]<\/p>\n\n\n\nProblem 2: Zeros: -3, -2, 1, 2; Degree: 4<\/h3>\n\n\n\n
Given zeros: ( -3, -2, 1, 2 ), the corresponding factors are:<\/p>\n\n\n\n
- \n
- ( (x + 3) ) for zero -3<\/li>\n\n\n\n
- ( (x + 2) ) for zero -2<\/li>\n\n\n\n
- ( (x – 1) ) for zero 1<\/li>\n\n\n\n
- ( (x – 2) ) for zero 2<\/li>\n<\/ul>\n\n\n\n
Thus, the polynomial is:<\/p>\n\n\n\n
[
f(x) = (x + 3)(x + 2)(x – 1)(x – 2)
]<\/p>\n\n\n\nTo expand this, we proceed step by step.<\/p>\n\n\n\n
- \n
- First, expand ( (x + 3)(x + 2) ):
[
(x + 3)(x + 2) = x^2 + 5x + 6
]<\/li>\n\n\n\n - Next, expand ( (x – 1)(x – 2) ):
[
(x – 1)(x – 2) = x^2 – 3x + 2
]<\/li>\n\n\n\n - Now multiply the two quadratics ( (x^2 + 5x + 6) ) and ( (x^2 – 3x + 2) ):
[
(x^2 + 5x + 6)(x^2 – 3x + 2) = x^4 + 2x^3 – 7x^2 + 4x + 12
]<\/li>\n<\/ol>\n\n\n\nThus, the polynomial is:
[
f(x) = x^4 + 2x^3 – 7x^2 + 4x + 12
]<\/p>\n\n\n\nProblem 3: Zeros: -1 (multiplicity 1), 4 (multiplicity 2); Degree: 3<\/h3>\n\n\n\n
Given zeros: ( -1 ) with multiplicity 1 and ( 4 ) with multiplicity 2, the corresponding factors are:<\/p>\n\n\n\n
- \n
- ( (x + 1) ) for zero -1<\/li>\n\n\n\n
- ( (x – 4)^2 ) for zero 4 with multiplicity 2<\/li>\n<\/ul>\n\n\n\n
Thus, the polynomial is:<\/p>\n\n\n\n
[
f(x) = (x + 1)(x – 4)^2
]<\/p>\n\n\n\nTo expand this expression:<\/p>\n\n\n\n
- \n
- First, expand ( (x – 4)^2 ):
[
(x – 4)^2 = x^2 – 8x + 16
]<\/li>\n\n\n\n - Now multiply ( (x + 1) ) by ( (x^2 – 8x + 16) ):
[
(x + 1)(x^2 – 8x + 16) = x^3 – 8x^2 + 16x + x^2 – 8x + 16
]<\/li>\n\n\n\n - Combine like terms:
[
f(x) = x^3 – 7x^2 + 8x + 16
]<\/li>\n<\/ol>\n\n\n\nSummary of Polynomials:<\/h3>\n\n\n\n
- \n
- ( f(x) = x^3 – 2x^2 – x + 2 )<\/li>\n\n\n\n
- ( f(x) = x^4 + 2x^3 – 7x^2 + 4x + 12 )<\/li>\n\n\n\n
- ( f(x) = x^3 – 7x^2 + 8x + 16 )<\/li>\n<\/ol>\n\n\n\n
Visualization of the Polynomials:<\/h3>\n\n\n\n
I’ll generate a graph that shows the roots and behavior of the polynomials.<\/p>\n\n\n\n
Here is a visual representation of the three polynomials based on their zeros:<\/p>\n\n\n\n
- \n
- The blue curve represents ( f(x) = x^3 – 2x^2 – x + 2 ) with zeros at -1, 1, and 2.<\/li>\n\n\n\n
- The red curve represents ( f(x) = x^4 + 2x^3 – 7x^2 + 4x + 12 ) with zeros at -3, -2, 1, and 2.<\/li>\n\n\n\n
- The green curve represents ( f(x) = x^3 – 7x^2 + 8x + 16 ) with zeros at -1 (multiplicity 1) and 4 (multiplicity 2).<\/li>\n<\/ul>\n\n\n\n
Each polynomial is plotted along with its zeros clearly marked, showing how the degree and multiplicity affect the shape of the curves. Let me know if you’d like more clarification!<\/p>\n\n\n\n
![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/02\/image-573-1024x650.png\")
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- First, expand ( (x – 4)^2 ):
- First, expand ( (x + 3)(x + 2) ):
- First, expand ( (x + 1)(x – 1) ) using the difference of squares formula: