# Solutions to graded problems

__Section 3.2__

2. Using long division:

So P(x) = (3x-2)(x^{2}-4)-12.

10. Write P(x) = 2x^{3}-10x^{2}-28x+60.
Then we are told that x =-3 is a solution of the

equation P (x) = 0, which means that (x + 3) must be a __factor__ of P(x), by
the factor

theorem. We use long division to factor P:

So now the original equation is equivalent to

2(x + 3)(x^{2}-8x + 10) = 0

(I pulled out a 2 for convenience.) To find the remaining two solutions of this
equation

(apart from x =-3), we can use the quadratic formula on the second bracket:

(x^{2}-8x + 10) = 0⇔⇔

So the equation has solutions x =-3, x = 4-and
x = 4 +.

16. We're told to use a graph to solve the equation x^{3}-3x^{2}-4x+12
= 0. This corresponds

to looking for x- intercepts of the graph y=x^{3}-3x^{2}-4x+12,
sketched below:

From here, we can immediately read o that the x-intercepts are at (-2,0), (2,0)

and (3,0). By the factor theorem , this means we can write x^{3}-3x^{2}-4x+12=

(x-3)(x-2)(x + 2). So the original equation has solutions x =-2, x = 2 and x =
3.

28. How do we solve a cubic equation such as this one?
First, collect all terms to one side:

x^{3} + 10x^{2} + 14x-60 = 0

From this point, our strategy is to inspect the graph and try to find at least one
x -

intercept with rational coordinates (this is what we will do in general, for a
polynomial

equation with degree≥3). The graph is sketched below:

On your calculators, you should use the calculate function to find the
x-intercepts

(carefully!) I have drawn a zoomed -in graph (below) to show what happens near

x =-6.

From your calculator, you should find that the x-intercepts are at x =-6, x≈

-5.741657...and x≈1.741657...Obviously, the latter two are not exact . To find

them exactly, we can factor our polynomial, using the information that x =-6 is
an

exact zero:

So x^{3}-3x^{2}-4x+12= 0 , (x+6)(x^{2} +4x-10) = 0 )⇒x
=-6 or x =,

So the final, exact solutions are x =-6, x =-2-and
x =-2 +

60. The information about the zeros gives that P(x) =
a(x-1)^{2}(x + 1)^{2}. a is a stretching

factor that will be determined by the other piece of information: P(2) = 4 , 4 =

a(1)^{2}(3)^{2} = 9a⇔a =.So
P(x) =.

62. The graph shows that the polynomial P has zeros at x
=-5, x = 2 and x = 5 (all

multiplicity one ), so we immediately have that P(x) = a(x-5)(x-2)(x + 5) (by the

factor theorem). To find the stretch factor a, we use one other point on the
graph: in

this case, the y-intercept (0,50): P(0) = 50⇔50 = a(-5)(-2)(5) = 50a⇔a =

1. So P(x) = (x-5)(x-2)(x + 5).

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