Arithmetic with Polynomials and Rational Expressions: Using Zeros of Polynomials to Construct Graphs (CCSS.A-APR.3)
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Common Core High School Algebra › Arithmetic with Polynomials and Rational Expressions: Using Zeros of Polynomials to Construct Graphs (CCSS.A-APR.3)
What are the zeros of the polynomial $f(x) = -2(x-3)(x+1)$?
$x=-3$ and $x=1$
$x=3$ and $x=-1$
$x=0,\ x=3,\ x=-1$
$x=2$ and $x=-1$
Explanation
Set each factor to zero: $x-3=0\Rightarrow x=3$ and $x+1=0\Rightarrow x=-1$. The leading coefficient $-2$ does not affect the zeros. A flips the signs. C adds an extra zero at $x=0$. D incorrectly treats the coefficient as a factor.
Which graph matches this polynomial $g(x) = (x+2)(x-1)^2$?
Crosses the $x$-axis at $x=-2$ and at $x=1$; falls to the left and rises to the right.
Touches the $x$-axis at $x=-2$ and crosses at $x=1$; falls to the left and rises to the right.
Crosses the $x$-axis at $x=-2$ and at $x=1$; rises to the left and falls to the right.
Crosses the $x$-axis at $x=-2$ and touches at $x=1$; falls to the left and rises to the right.
Explanation
Zeros come from factors: $x=-2$ (single, crosses) and $x=1$ (double, touches). Leading term is positive cubic, so left end down and right end up. D matches both the intercept behavior and end behavior. A crosses at $x=1$ (wrong multiplicity). B touches at $x=-2$ (wrong multiplicity). C has the wrong end behavior for a positive leading cubic.
What are the zeros of the polynomial $h(x) = 3(x+4)(x-2)^2$?
$x=-4$ and $x=2$
$x=4$ and $x=-2$
$x=-4,\ x=2,\ x=0$
$x=-\tfrac{4}{3}$ and $x=\tfrac{2}{3}$
Explanation
Set each factor to zero: $x+4=0\Rightarrow x=-4$ and $x-2=0\Rightarrow x=2$ (double root). The coefficient 3 does not create or move zeros. B flips signs. C adds a nonexistent zero at $x=0$. D incorrectly divides by the leading coefficient.
Which graph matches this polynomial $p(x) = - (x+3)(x+1)(x-2)$?
Crosses at $x=-3$, $x=-1$, and $x=2$; left end down and right end up.
Touches at $x=-1$ and crosses at $x=-3$ and $x=2$; left end up and right end down.
Crosses at $x=-3$, $x=-1$, and $x=2$; left end up and right end down.
Crosses at $x=-3$, $x=1$, and $x=2$; left end up and right end down.
Explanation
Zeros: $x=-3$, $x=-1$, and $x=2$, each from a single factor, so the graph crosses at all three. The leading coefficient is negative for an odd-degree polynomial, so left end up and right end down. A has the end behavior of a positive leading coefficient. B incorrectly shows a touch at $x=-1$. D uses the wrong sign for the zero from $(x+1)$.
What are the zeros of the polynomial $q(x) = 5x(x+5)$?
$x=5$ and $x=-5$
$x=0$ and $x=-5$
$x=0$ only
$x=-\tfrac{1}{5}$ and $x=-5$
Explanation
Set each factor to zero: $x=0$ from the factor $x$, and $x+5=0\Rightarrow x=-5$. The leading coefficient 5 does not affect the zeros. A mistakes $x=5$ for a zero. C omits the zero at $x=-5$. D incorrectly treats the coefficient as if it created a zero at $x=-\tfrac{1}{5}$.
What are the zeros of the polynomial $f(x)=(x-4)(x+1)$?
$x=-4,\ x=1$
$x=4,\ x=-1$
$x=4,\ x=1$
$x=-4,\ x=-1$
Explanation
Set each factor to zero: $x-4=0\Rightarrow x=4$ and $x+1=0\Rightarrow x=-1$. The graph would cross the $x$-axis at $x=4$ and $x=-1$. A and D flip one or both signs (confusing $x-a$ with $x+a$). C has the wrong sign for the second zero.
What are the zeros of the polynomial $g(x)=-2(x-3)^2$?
$x=-3$
$x=0,\ x=3$
$x=-2,\ x=3$
$x=3$
Explanation
Set the factor to zero: $x-3=0\Rightarrow x=3$. The coefficient $-2$ does not create or move zeros. Because the factor is squared, the graph would touch (bounce off) the $x$-axis at $x=3$. A uses the wrong sign. B invents an extra zero at $x=0$. C treats the leading coefficient as if it were a zero.
What are the zeros of the polynomial $h(x)=\tfrac{1}{2}(x+5)(x-1)(x-2)$?
$x=-5,\ x=1,\ x=2$
$x=5,\ x=-1,\ x=-2$
$x=-5,\ x=-1,\ x=2$
$x=1,\ x=2,\ x=\tfrac{1}{2}$
Explanation
Each factor gives a zero: $x+5=0\Rightarrow x=-5$, $x-1=0\Rightarrow x=1$, $x-2=0\Rightarrow x=2$. The coefficient $\tfrac{1}{2}$ does not affect zeros. The graph would cross the $x$-axis at all three zeros. B flips all signs. C flips the sign on $x-1$. D incorrectly treats the coefficient as a zero.
What are the zeros of the polynomial $p(x)=x(x+2)^2$?
$x=2$
$x=0,\ x=2$
$x=0,\ x=-2$
$x=-2$
Explanation
Set each factor to zero: $x=0$ and $x+2=0\Rightarrow x=-2$. The $(x+2)^2$ gives an even multiplicity, so the graph would touch the $x$-axis at $x=-2$ and cross at $x=0$. A and D list only one of the zeros. B has the wrong sign for the double root.
What are the zeros of the polynomial $q(x)=-(x+4)(x-6)$?
$x=-4,\ x=6$
$x=4,\ x=6$
$x=-4,\ x=-6$
$x=4,\ x=-6$
Explanation
Set each factor to zero: $x+4=0\Rightarrow x=-4$ and $x-6=0\Rightarrow x=6$. The negative leading coefficient does not change the zeros; it only flips the parabola's opening. B and D use the wrong sign for one factor. C flips the sign on the second factor.