Problem 1:
The graph of a cubic polynomial
y = a x 3 + b x 2 + c x + d
is shown below. Find the coefficients a, b, c and d.
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Solution to Problem 1:
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This polynomial has a zero of multiplicity 1 at x = -2 and a zero of multiplicity 2 at x = 1. Hence the polynomial may be written as
y = a (x + 2)(x - 1) 2
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This polynomial has a y intercept (0 , 1). Hence
1 = a (0 + 2)(0 - 1) 2
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Solve for a to obtain
a = 1 / 2
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The polynomial may now be written as follows
y = (1 / 2) (x + 2)(x - 1) 2
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Expand to obtain
y = (1 / 2) x 3 -(3 / 2) x + 1
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We now identify the coefficients as follows
a = 1/2 , b = 0, c = -3/2 and d = 1
Problem 2:
The graph of the polynomial
y = a x 4 + b x 3 + c x 2 + d x + e
is shown below. Find the coefficients a, b, c, d and e.
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Solution to Problem 2:
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This polynomial has a zero of multiplicity 2 at x = - 2 and a zero of multiplicity 2 at x = 2. Hence it may be written as
y = a (x + 2) 2 (x - 2) 2
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We now use the y intercept at (0 , -2) to write the equation
-2 = a (0 + 2) 2 (0 - 2) 2
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Solve the above for a to obtain
a = - 1 / 8
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We now write the polynomial as follows
y = (-1 / 8) (x + 2) 2 (x - 2) 2
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Expand
y = (-1 / 8) x 4 + x 2 - 2
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We now identify the coefficients
a = -1 / 8, b = 0, c = 1, d = 0, e = -2
Problem 3:
The polynomial
f(x) = x 6 + 4 x 5 + x 4 - 12 x 3 - 11 x 2 + 4 x + 4
has a zero of multiplicity 2 at x = - 2. Find the other real zeros.
Solution to Problem 3:
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If f has a zero of multiplicity 2, then it may be written as follows
f(x) = (x + 2) 2 Q(x)
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Where Q(x) is a polynomial of degree 4 and may be found by division
Q(x) = f(x) / (x + 2) 2 = x 4 -3 x 2 + 1
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Polynomial f may now be written as
f(x) = (x + 2) 2 (x 4 -3 x 2 + 1)
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The remaining zeros of polynomial f may be found by solving the equation
x 4 -3 x 2 + 1 = 0
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It is an equation of the quadratic type with solutions
( √(5) + 1 ) / 2 , ( √(5) - 1 ) / 2 , ( - √(5) - 1 ) / 2 , ( - √(5) + 1 ) / 2
Problem 4:
The polynomial
f(x) = 3 x 4 + 5 x 3 - 17 x 2 - 25 x + 10
has irrational zeros at + √(5) and - √(5). Find the other zeros.
Solution to Problem 4:
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polynomial f may be written as
f(x) = (x + √(5)) (x - √(5)) Q(x) = (x 2 - 5) Q(x)
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Q(x) may be found by division
Q(x) = f(x) / (x 2 - 5) = 3 x 2 + 5 x - 2
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Hence f(x) may be written as
f(x) = (x 2 - 5) (3 x 2 + 5 x - 2 )
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The remaining zeros may found by solving the equation
3 x 2 + 5 x - 2 = 0
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Solve the above equation to find the remaining zeros of f.
-2 and 1 / 3
Problem 5:
A polynomial of degree 4 has a positive leading coefficient and simple zeros (i.e. zeros of multiplicity 1) at x = 2, x = - 2, x = 1 and x = -1. Is the y intercept of the graph of this polynomial positive or negative?
Solution to Problem 5:
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All polynomials with degree 4 and positive leading coefficient will have a graph that rises to the left and to the right. And since the polynomial has two negative zeros and two positive zeros, then the only possibility for the y intercept is to be positive.
Problem 6:
The graph of polynomial p is shown below.
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a - Is the degree of p even or odd?
b - Is the leading coefficient negative or positive?
c - Can you find the degree from the graph of p? Explain.
Solution to Problem 6:
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a - odd
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b - negative
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c - No. The degree of a polynomial depends on the real and complex zeros. The graph shows only the real zeros. Hence, not enough information is given to find the degree of the polynomial.
Problem 7:
Give 4 different reasons why the graph below cannot be the graph of the polynomial p give by.
p(x) = x 4 - x 2 + 1
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Solution to Problem 7:
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1 - The given polynomial has degree 4 and positive leading coefficient and the graph should rise on the left and right sides.
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2 - p(0) = 1, graph shows negative value.
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3 - the equation x 4 - x 2 + 1 = 0 has no solution which suggests that the polynomial p(x) = x 4 - x 2 + 1 has no zeros. The graph shows x intercepts.
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4 - The graph has a zero of multiplicity 2, a zero of multiplicity 3 and and a zero of multiplicity 1. So the degree of the graphed polynomial should be at least 6. The given polynomial has degree 4.
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