

Give four different reasons why the graph below cannot possibly be the graph of the polynomial function \( p(x) = x^4x^2+1 \).
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Solution
The four reasons are:
1) The given polynomial function is even and therefore its graph must be symmetric with respect to the y axis. The given graph is not symmetric with respect to the y axis.
2) The given polynomial function does not have real zeros (discriminant = 3 : negative). The given graph has x intercepts which must corresponds to real zeros.
3) The y intercept calculated using p(x)( p(0) = 0^40^2+1 = 1) is positive. The y intercept of the graph is negative.
4) Having a leading coefficient ( = 1) positive and an even degree ( = 4), the polynomial must have a graph with the right and left both rising. In the given graph, they are both falling.

Match the polynomial functions to their graphs where all x intercepts are shown.
$$f(x) = (x+1)(x1)^2(x+2)^2$$
$$g(x) = (x+1)(x1)^4$$
$$h(x) = (x+1)(x1)^3(x3)$$
$$i(x) = (x+1)^2(x2)^3$$
$$j(x) = (x+1)^2(1x)(x2)^2$$
$$k(x) =(x+1)^2(x1)^2(x3)$$
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Solution
According to their equations, all 6 given polynomial functions are of degree 5. However their leading coefficients are of different signs. We classify the 6 polynomials into 2 groups: I and II
Group I  Given polynomials with positive leading coefficients
f(x) = (x + 1)(x  1)^{2}(x + 2)^{2}
h(x) = (x + 1)(x  1)^{3}(x  3)
i(x) = (x + 1)^{2}(x  2)^{3}
Having degree 5 (odd) and leading coefficients positive, each of the graphs of the above polynomials (f, h and i) has the following graphical properties:
as x ^{____}> ∞ , y ^{____}> ∞ (the right hand side of the graphs rises)
as x ^{____}>  ∞ y , ^{____}>  ∞ (the left hand side of the graph falls)
The given graphs in parts a) c) and e) have the above properties with different x intercepts and their multiplicities. Hence
Polynomial f(x) = (x + 1)(x  1)^{2}(x + 2)^{2} has a zero of multiplicity 1 at x = 1 , a zero of multiplicity 2 at x = 1 and a zero of multiplicity 2 at x =  2 and should correspond to the graph in part e).
Polynomial h(x) = (x + 1)(x  1)^{3}(x  3) has a zero of multiplicity 1 at x = 1 , a zero of multiplicity 3 at x = 1 and a zero of multiplicity 1 at x = 3 and should correspond to the graph in part a).
Polynomial i(x) = (x + 1)^{2}(x  2)^{3} has a zero of multiplicity 2 at x = 1 and a zero of multiplicity 3 at x = 2 and should correspond to the graph in part c).
Group II  Given polynomials with negative leading coefficients
The polynomial functions g, j and k, when expanded, have leading coefficients that are negative.
g(x) =  (x + 1)(x  1)^{4}
j(x) = (x + 1)^{2}(1  x)(x  2)^{2}
k(x) =  (x + 1)^{2}(x  1)^{2}(x  3)
Having degree 5 (odd) and leading coefficients negative, each of the graphs of the above polynomials (g, j and k) has the following graphical properties:
as x ^{____}> ∞ , y ^{____}>  ∞ (the right hand side of the graphs falls)
as x ^{____}>  ∞ y , ^{____}> ∞ (the left hand side of the graph rises)
The given graphs in parts b) d) and f) have the above properties with different x intercepts and their multiplicities. Hence
Polynomial g(x) =  (x + 1)(x  1)^{4} has a zero of multiplicity 1 at x = 1 , a zero of multiplicity 4 at x = 1 and should correspond to the graph in part f).
Polynomial j(x) = (x + 1)^{2}(1  x)(x  2)^{2} has a zero of multiplicity 2 at x = 1 , a zero of multiplicity 1 at x = 1 and a zero of multiplicity 2 at x = 2 and should correspond to the graph in part d).
Polynomial k(x) =  (x + 1)^{2}(x  1)^{2}(x  3) has a zero of multiplicity 2 at x = 1 , a zero of multiplicity 2 at x = 1 and a zero of multiplicity 1 at x = 3 and should correspond to the graph in part b).

