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SAT II Math II : SAT Subject Test in Math II

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Example Questions

Example Question #531 : Sat Subject Test In Math Ii

Give the set of all real solutions of the equation $2x -11x^{\frac{1}{2}}+15 = 0$ $\displaystyle 2x -11x^{\frac{1}{2}}+15 = 0$.

Possible Answers:

$\left \{ - 6\frac{1}{4}, -9 \right \}$ $\displaystyle \left \{ - 6\frac{1}{4}, -9 \right \}$

$\left \{ - 6\frac{1}{4} , 6\frac{1}{4} \right \}$ $\displaystyle \left \{ - 6\frac{1}{4} , 6\frac{1}{4} \right \}$

$\left \{ 6\frac{1}{4}, 9 \right \}$ $\displaystyle \left \{ 6\frac{1}{4}, 9 \right \}$

The equation has no real solutions.

$\left \{-9, 9 \right \}$ $\displaystyle \left \{-9, 9 \right \}$

Correct answer:

$\left \{ 6\frac{1}{4}, 9 \right \}$ $\displaystyle \left \{ 6\frac{1}{4}, 9 \right \}$

Explanation:

Set $y = x^{\frac{1}{2}}$ $\displaystyle y = x^{\frac{1}{2}}$. Then $x = \left (x^{\frac{1}{2}} \right )^{2} = y ^{2}$ $\displaystyle x = \left (x^{\frac{1}{2}} \right )^{2} = y ^{2}$.

$2x -11x^{\frac{1}{2}}+15 = 0$ $\displaystyle 2x -11x^{\frac{1}{2}}+15 = 0$ can be rewritten as

$2\left (x^{\frac{1}{2}} \right )^{2} -11 \left (x^{\frac{1}{2}} \right )+15 = 0$ $\displaystyle 2\left (x^{\frac{1}{2}} \right )^{2} -11 \left (x^{\frac{1}{2}} \right )+15 = 0$

Substituting $y^{2}$ $\displaystyle y^{2}$ for $\displaystyle x$ and $\displaystyle y$ for $x^{\frac{1}{2}}$ $\displaystyle x^{\frac{1}{2}}$, the equation becomes

$2y ^{2} -11 y +15 = 0$ $\displaystyle 2y ^{2} -11 y +15 = 0$

a quadratic equation in $\displaystyle y$.

This can be solved using the $\displaystyle ac$ method. We are looking for two integers whose sum is $\displaystyle -11$ and whose product is $2\cdot 15 = 30$ $\displaystyle 2\cdot 15 = 30$. Through some trial and error, the integers are found to be $\displaystyle -6$ and $\displaystyle -5$, so the above equation can be rewritten, and solved using grouping, as

$2y ^{2} -6y -5y +15 = 0$ $\displaystyle 2y ^{2} -6y -5y +15 = 0$

$(2y ^{2} -6y) - (5y-15 )= 0$ $\displaystyle (2y ^{2} -6y) - (5y-15 )= 0$

$\displaystyle 2y (y-3) -5(y-3) = 0$

$\displaystyle (2y -5) (y-3) = 0$

By the Zero Product Principle, one of these factors is equal to zero:

Either:

$\displaystyle 2y-5 = 0$

$\displaystyle 2y - 5 + 5 = 0 + 5$

2y = 5 $\displaystyle 2y = 5$

$\frac{2y }{2}= \frac{5}{2}$ $\displaystyle \frac{2y }{2}= \frac{5}{2}$

$y= \frac{5}{2}$ $\displaystyle y= \frac{5}{2}$

Substituting back:

$x^{\frac{1}{2}} = \frac{5}{2}$ $\displaystyle x^{\frac{1}{2}} = \frac{5}{2}$, or $\sqrt{x} = \frac{5}{2}$ $\displaystyle \sqrt{x} = \frac{5}{2}$

$(\sqrt{x} )^{2}= \left (\frac{5}{2} \right )^{2}$ $\displaystyle (\sqrt{x} )^{2}= \left (\frac{5}{2} \right )^{2}$

$x = \frac{25}{4} = 6\frac{1}{4}$ $\displaystyle x = \frac{25}{4} = 6\frac{1}{4}$

Or:

$\displaystyle y - 3 = 0$

$\displaystyle y - 3 + 3 = 0 + 3$

y = 3 $\displaystyle y = 3$

Substituting back:

$x^{\frac{1}{2}} =3$ $\displaystyle x^{\frac{1}{2}} =3$, or $\sqrt{x} = 3$ $\displaystyle \sqrt{x} = 3$

$(\sqrt{x} )^{2}= 3^{2}$ $\displaystyle (\sqrt{x} )^{2}= 3^{2}$

x =9 $\displaystyle x =9$

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Example Question #82 : Single Variable Algebra

Define a function $f(x) = x^{2} - \cos 4x$ $\displaystyle f(x) = x^{2} - \cos 4x$.

$\displaystyle f(c) = 3$ for exactly one positive value of $\displaystyle c$; this is on the interval (0,5) $\displaystyle (0,5)$. Which of the following is true of $\displaystyle c$?

Possible Answers:

$c \in (1, 2)$ $\displaystyle c \in (1, 2)$

$c \in (0, 1)$ $\displaystyle c \in (0, 1)$

$c \in (2, 3)$ $\displaystyle c \in (2, 3)$

$c \in (4,5 )$ $\displaystyle c \in (4,5 )$

$c \in (3, 4)$ $\displaystyle c \in (3, 4)$

Correct answer:

$c \in (1, 2)$ $\displaystyle c \in (1, 2)$

Explanation:

Define $g(x)= f(x) - 3 = x^{2} - \cos 4x- 3$ $\displaystyle g(x)= f(x) - 3 = x^{2} - \cos 4x- 3$. Then, if $\displaystyle g(c)= f(c) - 3= 0$, it follows that $\displaystyle f(c) = 3$.

By the Intermediate Value Theorem (IVT), if g(x) $\displaystyle g(x)$ is a continuous function, and g(a) $\displaystyle g(a)$ and g(b) $\displaystyle g(b)$ are of unlike sign, then $\displaystyle g(c) = 0$ for some $c \in (a, b)$ $\displaystyle c \in (a, b)$. $\displaystyle 5x$ and $\cos 2x$ $\displaystyle \cos 2x$ are both continuous everywhere, so g(x) $\displaystyle g(x)$ is a continuous function, so the IVT applies here.

Evaluate g(x) $\displaystyle g(x)$ for each of the following values: $\left \{ 0, 1, 2, 3, 4, 5 \right \}$ $\displaystyle \left \{ 0, 1, 2, 3, 4, 5 \right \}$:

$g(x) = x^{2} - \cos 4x- 3$ $\displaystyle g(x) = x^{2} - \cos 4x- 3$

$g(0) =0^{2} - \cos 4 (0)- 3$ $\displaystyle g(0) =0^{2} - \cos 4 (0)- 3$

$=0 - \cos 0- 3$ $\displaystyle =0 - \cos 0- 3$

$\displaystyle =0 -1- 3$

= -4 $\displaystyle = -4$

$g(1) =1^{2} - \cos 4 (1)- 3$ $\displaystyle g(1) =1^{2} - \cos 4 (1)- 3$

$=1 - \cos 4- 3$ $\displaystyle =1 - \cos 4- 3$

$\approx 1 - ( -0.65 ) - 3$ $\displaystyle \approx 1 - ( -0.65 ) - 3$

$\approx -1.35$ $\displaystyle \approx -1.35$

$g(2) =2^{2} - \cos 4 (2)- 3$ $\displaystyle g(2) =2^{2} - \cos 4 (2)- 3$

$\approx 4 - ( -0.15) - 3$ $\displaystyle \approx 4 - ( -0.15) - 3$

$\approx 1.15$ $\displaystyle \approx 1.15$

$g(3) =3^{2} - \cos 4 (3)- 3$ $\displaystyle g(3) =3^{2} - \cos 4 (3)- 3$

$\approx 9- 0.84 - 3$ $\displaystyle \approx 9- 0.84 - 3$

$\approx 5.16$ $\displaystyle \approx 5.16$

$g(4) =4^{2} - \cos 4 (4)- 3$ $\displaystyle g(4) =4^{2} - \cos 4 (4)- 3$

$\approx 16 - (-0.96 )- 3$ $\displaystyle \approx 16 - (-0.96 )- 3$

$\approx 13.96$ $\displaystyle \approx 13.96$

$g(5) =5^{2} - \cos 4 (5)- 3$ $\displaystyle g(5) =5^{2} - \cos 4 (5)- 3$

$\approx 25 - 0.41 - 3$ $\displaystyle \approx 25 - 0.41 - 3$

$\approx 21.59$ $\displaystyle \approx 21.59$

Only in the case of (1, 2) $\displaystyle (1, 2)$ does it hold that g (x) $\displaystyle g (x)$ assumes a different sign at the endpoints - $\displaystyle g(1) < 0 < g(2)$. By the IVT, $\displaystyle g(c) = 0$, and $\displaystyle f(c) = 3$, for some $c \in (1, 2)$ $\displaystyle c \in (1, 2)$.

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Example Question #83 : Single Variable Algebra

Which of the following is a cube root of $\displaystyle -64$?

Possible Answers:

$2+ 2 \sqrt{3}$ $\displaystyle 2+ 2 \sqrt{3}$

$2 + 2 i \sqrt{5}$ $\displaystyle 2 + 2 i \sqrt{5}$

$2+ 2 i \sqrt{3}$ $\displaystyle 2+ 2 i \sqrt{3}$

$2 + 2 \sqrt{5}$ $\displaystyle 2 + 2 \sqrt{5}$

None of the other choices gives a correct response.

Correct answer:

$2+ 2 i \sqrt{3}$ $\displaystyle 2+ 2 i \sqrt{3}$

Explanation:

Let $\displaystyle x$ be a cube root of $\displaystyle -64$. The question is to find a solution of the equation

$x^{3} =- 64$ $\displaystyle x^{3} =- 64$.

One way to solve this is to add 64 to both sides:

$x^{3} + 64 =- 64 + 64$ $\displaystyle x^{3} + 64 =- 64 + 64$

$x^{3} + 64 =0$ $\displaystyle x^{3} + 64 =0$

64 is a perfect cube, so, as the sum of cubes, the left expression can be factored:

$x^{3} + 4 ^{3} =0$ $\displaystyle x^{3} + 4 ^{3} =0$

$(x+ 4 )(x^{2} - x \cdot 4 + 4 ^{2}) = 0$ $\displaystyle (x+ 4 )(x^{2} - x \cdot 4 + 4 ^{2}) = 0$

$(x+ 4 )(x^{2} - 4x + 16) = 0$ $\displaystyle (x+ 4 )(x^{2} - 4x + 16) = 0$

We can set both factors equal to zero and solve:

$\displaystyle x + 4 = 0$

x =- 4 $\displaystyle x =- 4$

$\displaystyle -4$ is a cube root of $\displaystyle -64$; however, this is not one of the choices.

Setting

$x^{2} - 4x + 16 = 0$ $\displaystyle x^{2} - 4x + 16 = 0$,

we can make use of the quadratic formula, setting $\displaystyle a = 1, b= -4, c= 16$ in the following:

$x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ $\displaystyle x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

$x = \frac{-(-4) \pm \sqrt{(-4)^{2}-4(1)(16)}}{2(1)}$ $\displaystyle x = \frac{-(-4) \pm \sqrt{(-4)^{2}-4(1)(16)}}{2(1)}$

$x = \frac{4 \pm \sqrt{16-64}}{2}$ $\displaystyle x = \frac{4 \pm \sqrt{16-64}}{2}$

$x = \frac{4 \pm \sqrt{-48 }}{2}$ $\displaystyle x = \frac{4 \pm \sqrt{-48 }}{2}$

$x = \frac{4 \pm i \sqrt{48 }}{2}$ $\displaystyle x = \frac{4 \pm i \sqrt{48 }}{2}$

$x = \frac{4 \pm i \sqrt{16} \cdot \sqrt{3}}{2}$ $\displaystyle x = \frac{4 \pm i \sqrt{16} \cdot \sqrt{3}}{2}$

$x = \frac{4 \pm 4 i \sqrt{3}}{2}$ $\displaystyle x = \frac{4 \pm 4 i \sqrt{3}}{2}$

$x = 2 \pm 2 i \sqrt{3}$ $\displaystyle x = 2 \pm 2 i \sqrt{3}$

$2 - 2 i \sqrt{3}$ $\displaystyle 2 - 2 i \sqrt{3}$ and $2 + 2 i \sqrt{3}$ $\displaystyle 2 + 2 i \sqrt{3}$ are both cube roots of $\displaystyle -64$; $2 - 2 i \sqrt{3}$ $\displaystyle 2 - 2 i \sqrt{3}$ is not a choice, but $2 + 2 i \sqrt{3}$ $\displaystyle 2 + 2 i \sqrt{3}$ is.

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Example Question #84 : Single Variable Algebra

A polynomial of degree 4 has as its lead term $x^{4}$ $\displaystyle x^{4}$ and has rational coefficients. Two of its zeroes are 4+i $\displaystyle 4+i$ and 5+i $\displaystyle 5+i$ What is this polynomial?

Possible Answers:

$P(x)= x^{4} +18x^{3} + 129x^{2}+342 x+360$ $\displaystyle P(x)= x^{4} +18x^{3} + 129x^{2}+342 x+360$

$P(x)= x^{4} - 18x^{3} + 123x^{2} - 378x+442$ $\displaystyle P(x)= x^{4} - 18x^{3} + 123x^{2} - 378x+442$

$P(x)= x^{4} +18x^{3} + 123x^{2}+378x+442$ $\displaystyle P(x)= x^{4} +18x^{3} + 123x^{2}+378x+442$

Insufficient information exists to determine the polynomial.

$P(x)= x^{4} - 18x^{3} + 129x^{2}-342 x+360$ $\displaystyle P(x)= x^{4} - 18x^{3} + 129x^{2}-342 x+360$

Correct answer:

$P(x)= x^{4} - 18x^{3} + 123x^{2} - 378x+442$ $\displaystyle P(x)= x^{4} - 18x^{3} + 123x^{2} - 378x+442$

Explanation:

A fourth-degree, or quartic, polynomial has four zeroes, if a zero of multiplicity $\displaystyle n$ is counted $\displaystyle n$ times. Since its lead term is $x^{4}$ $\displaystyle x^{4}$, we know that

$p(x) = (x-b_{1}) (x-b_{2}) (x-b_{3})(x-b_{4})$ $\displaystyle p(x) = (x-b_{1}) (x-b_{2}) (x-b_{3})(x-b_{4})$

A polynomial with rational coefficients has its imaginary zeroes in conjugate pairs. Since p(x) $\displaystyle p(x)$ is such a polynomial, then, since 4+i $\displaystyle 4+i$ is a zero, so is its complex conjugate 4 -i $\displaystyle 4 -i$; similarly, since 5+i $\displaystyle 5+i$ is a zero, so is its complex conjugate 5-i $\displaystyle 5-i$. Substituting these four values for the four $b_{n}$ $\displaystyle b_{n}$ values:

$\displaystyle p(x) = [x- (4+i)] [x-(4-i)] [x- (5+i)] [x-(5-i)]$

This can be rewritten as

$\displaystyle p(x) = [x- 4- i] [x- 4 + i] [x- 5 - i] [x- 5+ i]$

$\displaystyle p(x) = [(x- 4) - i] [(x- 4) + i] [(x- 5) - i] [(x- 5) + i]$

Multiply the first two factors using the difference of squares pattern, then the square of a binomial pattern:

$\displaystyle [(x- 4) - i] [(x- 4) + i]$

$= (x-4)^{2} - i^{2}$ $\displaystyle = (x-4)^{2} - i^{2}$

$= x^{2} -2 \cdot x \cdot 4 + 4^{2} - (-1)$ $\displaystyle = x^{2} -2 \cdot x \cdot 4 + 4^{2} - (-1)$

$= x^{2} -8x + 16 +1$ $\displaystyle = x^{2} -8x + 16 +1$

$= x^{2} -8x + 17$ $\displaystyle = x^{2} -8x + 17$

Multiply the last two factors similarly:

$\displaystyle [(x- 5) - i] [(x- 5) + i]$

$= (x-5)^{2} - i^{2}$ $\displaystyle = (x-5)^{2} - i^{2}$

$= x^{2} -2 \cdot x \cdot 5 + 5^{2} - (-1)$ $\displaystyle = x^{2} -2 \cdot x \cdot 5 + 5^{2} - (-1)$

$= x^{2} -10x + 25+1$ $\displaystyle = x^{2} -10x + 25+1$

$= x^{2} -10x + 26$ $\displaystyle = x^{2} -10x + 26$

Thus,

$p(x)= ( x^{2} -8x + 17) ( x^{2} -10x + 26)$ $\displaystyle p(x)= ( x^{2} -8x + 17) ( x^{2} -10x + 26)$

Multiply:

$x^{2} -8x + 17$ $\displaystyle x^{2} -8x + 17$

$\underline{ x^{2} -10x + 26}$ $\displaystyle \underline{ x^{2} -10x + 26}$

$26 x^{2} -208 x +442$ $\displaystyle 26 x^{2} -208 x +442$

$-10 x^{3} +80x^{2} -170x$ $\displaystyle -10 x^{3} +80x^{2} -170x$

$\underline{x^{4} -8x^{3} +17x^{2}}$ $\displaystyle \underline{x^{4} -8x^{3} +17x^{2}}$________________

$x^{4} - 18x^{3} + 123x^{2} - 378x+442$ $\displaystyle x^{4} - 18x^{3} + 123x^{2} - 378x+442$.

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Example Question #531 : Sat Subject Test In Math Ii

Define a function $p (x) = x ^{6} - 7x^{4}$ $\displaystyle p (x) = x ^{6} - 7x^{4}$.

$\displaystyle p (c) = 5$ for exactly one value of $\displaystyle c$ on the interval (0, 5) $\displaystyle (0, 5)$. Which statement is true about $\displaystyle c$ ?

Possible Answers:

$c \in (3,4)$ $\displaystyle c \in (3,4)$

$c \in (0.1)$ $\displaystyle c \in (0.1)$

$c \in (2,3)$ $\displaystyle c \in (2,3)$

$c \in (1,2)$ $\displaystyle c \in (1,2)$

$c \in (4,5)$ $\displaystyle c \in (4,5)$

Correct answer:

$c \in (2,3)$ $\displaystyle c \in (2,3)$

Explanation:

Define $g(x) = p(x) -5 = x ^{6} - 7x^{4} -5$ $\displaystyle g(x) = p(x) -5 = x ^{6} - 7x^{4} -5$. Then, if $\displaystyle g(c) = p(c) -5 = 0$, it follows that $\displaystyle p(c) = 5$.

By the Intermediate Value Theorem (IVT), if g(x) $\displaystyle g(x)$ is a continuous function, and g(a) $\displaystyle g(a)$ and g(b) $\displaystyle g(b)$ are of unlike sign, then $\displaystyle g(c) = 0$ for some $c \in (a, b)$ $\displaystyle c \in (a, b)$. As a polynomial, g(x) $\displaystyle g(x)$ is a continuous function, so the IVT applies here.

Evaluate g(x) $\displaystyle g(x)$ for each of the following values: $\left \{ 0, 1, 2, 3, 4, 5 \right \}$ $\displaystyle \left \{ 0, 1, 2, 3, 4, 5 \right \}$:

$g(x) = x ^{6} - 7x^{4} -5$ $\displaystyle g(x) = x ^{6} - 7x^{4} -5$

$g(0) = 0 ^{6} - 7 \cdot 0 ^{4} -5$ $\displaystyle g(0) = 0 ^{6} - 7 \cdot 0 ^{4} -5$

$\displaystyle = 0 - 0 - 5$

= -5 $\displaystyle = -5$

$g(1) = 1 ^{6} - 7 \cdot 1 ^{4} -5$ $\displaystyle g(1) = 1 ^{6} - 7 \cdot 1 ^{4} -5$

$\displaystyle = 1 - 7 - 5$

= -11 $\displaystyle = -11$

$g(2) = 2 ^{6} - 7 \cdot 2 ^{4} -5$ $\displaystyle g(2) = 2 ^{6} - 7 \cdot 2 ^{4} -5$

$=64 - 7 \cdot 16 -5$ $\displaystyle =64 - 7 \cdot 16 -5$

$\displaystyle =64 - 112 -5$

= -53 $\displaystyle = -53$

$g(3) =3 ^{6} - 7 \cdot 3 ^{4} -5$ $\displaystyle g(3) =3 ^{6} - 7 \cdot 3 ^{4} -5$

$=729 - 7 \cdot 81 -5$ $\displaystyle =729 - 7 \cdot 81 -5$

$\displaystyle =729 - 567-5$

=157 $\displaystyle =157$

$g(4) =4 ^{6} - 7 \cdot 4 ^{4} -5$ $\displaystyle g(4) =4 ^{6} - 7 \cdot 4 ^{4} -5$

$=4,096 - 7 \cdot 256 -5$ $\displaystyle =4,096 - 7 \cdot 256 -5$

$\displaystyle =4,096 -1,792 -5$

=2,299 $\displaystyle =2,299$

$g(5) =5 ^{6} - 7 \cdot 5 ^{4} -5$ $\displaystyle g(5) =5 ^{6} - 7 \cdot 5 ^{4} -5$

$= 15,625 - 7 \cdot 625 -5$ $\displaystyle = 15,625 - 7 \cdot 625 -5$

$\displaystyle = 15,625 - 4,375 -5$

$\displaystyle = 11,245$

Only in the case of (2,3) $\displaystyle (2,3)$ does it hold that g (x) $\displaystyle g (x)$ assumes different signs at the endpoints - $\displaystyle g(2) < 0 < g(3)$. By the IVT, $\displaystyle g(c) = 0$, and $\displaystyle f(c) = 5$, for some $c \in (2,3)$ $\displaystyle c \in (2,3)$.

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Example Question #81 : Single Variable Algebra

What is a possible root to $\displaystyle y=2x^2-9x-5$?

Possible Answers:

$\displaystyle -5$

$-\frac{3}{2}$ $\displaystyle -\frac{3}{2}$

$\frac{2}{9}$ $\displaystyle \frac{2}{9}$

$-\frac{1}{2}$ $\displaystyle -\frac{1}{2}$

$\frac{1}{2}$ $\displaystyle \frac{1}{2}$

Correct answer:

$-\frac{1}{2}$ $\displaystyle -\frac{1}{2}$

Explanation:

Factor the trinomial.

The multiples of the first term is 1,2 $\displaystyle 1,2$.

The multiples of the third term is 1,5 $\displaystyle 1,5$.

We can then factor using these terms.

$\displaystyle y=(2x+1)(x-5)$

Set the equation to zero.

$\displaystyle 0=(2x+1)(x-5)$

This means that each product will equal zero.

$\displaystyle 2x+1 = 0 , x-5 = 0$

The roots are either $x=-\frac{1}{2} , -5$ $\displaystyle x=-\frac{1}{2} , -5$

The answer is: $-\frac{1}{2}$ $\displaystyle -\frac{1}{2}$

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Example Question #11 : Factoring And Finding Roots

Which of the following could be a solution for the equation $\displaystyle x^2-17x+30=0$?

Possible Answers:

x=-15 $\displaystyle x=-15$

x=2 $\displaystyle x=2$

x=-2 $\displaystyle x=-2$

There are no solutions for the equation.

Correct answer:

x=2 $\displaystyle x=2$

Explanation:

From the discriminant, b^2-4ac $\displaystyle b^2-4ac$, we know that this equation will have two solutions:

$\displaystyle (-17)^2-4(1)(30)>0$

Next, factor the equation $\displaystyle x^2-17x+30=0$.

$\displaystyle (x-5)(x-12)=0$

Finally, solve for $\displaystyle x$.

$x=5\text{ and }x=12$ $\displaystyle x=5\text{ and }x=12$

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Example Question #81 : Single Variable Algebra

Which of the following polynomials has $x^{2} - 1$ $\displaystyle x^{2} - 1$ as a factor?

Possible Answers:

$x^{2} +13x +15 x ^{2}+ 17 +14$ $\displaystyle x^{2} +13x +15 x ^{2}+ 17 +14$

$x^{4} - 13x ^{3} + 7x ^{2}+ 14x - 9$ $\displaystyle x^{4} - 13x ^{3} + 7x ^{2}+ 14x - 9$

$x^{4} -11x ^{3} + 6x^{2} + 11x - 7$ $\displaystyle x^{4} -11x ^{3} + 6x^{2} + 11x - 7$

None of these

$x^{4} + 5x^{2} - 6x - 7x + 5$ $\displaystyle x^{4} + 5x^{2} - 6x - 7x + 5$

Correct answer:

$x^{4} -11x ^{3} + 6x^{2} + 11x - 7$ $\displaystyle x^{4} -11x ^{3} + 6x^{2} + 11x - 7$

Explanation:

One way to work this problem is as follows:

Factor $x^{2} - 1$ $\displaystyle x^{2} - 1$ using the difference of squares pattern:

$x^{2} - 1^{2} = (x+1)(x-1)$ $\displaystyle x^{2} - 1^{2} = (x+1)(x-1)$

Consequently, any polynomial divisible by $x^{2} - 1$ $\displaystyle x^{2} - 1$ must be divisible by both x+1 $\displaystyle x+1$ and x - 1 $\displaystyle x - 1$.

A polynomial is divisible by x - 1 $\displaystyle x - 1$ if and only if the sum of its coefficients is 0. Add the coefficients for each given polynomial:

$x^{2} +13x +15 x ^{2}+ 17 +14$ $\displaystyle x^{2} +13x +15 x ^{2}+ 17 +14$

$\displaystyle 1+13+15 +17+14 = 60$

$x^{4} + 5x^{2} - 6x - 7x + 5$ $\displaystyle x^{4} + 5x^{2} - 6x - 7x + 5$:

$\displaystyle 1+5 - 6 -7 +5 = -2$

$x^{4} - 13x ^{3} + 7x ^{2}+ 14x - 9$ $\displaystyle x^{4} - 13x ^{3} + 7x ^{2}+ 14x - 9$:

$\displaystyle 1 -13 +7 +14 - 9 = 0$

$x^{4} -11x ^{3} + 6x^{2} + 11x - 7$ $\displaystyle x^{4} -11x ^{3} + 6x^{2} + 11x - 7$:

$\displaystyle 1 - 11 + 6 + 11 - 7 = 0$

The last two polynomials are both divisible by x - 1 $\displaystyle x - 1$. The other two can be eliminated as correct choices.

A polynomial is divisible by x+1 $\displaystyle x+1$ if and only if the alternating sum of its coefficients is 0- that is, if every other coefficient is reversed in sign and the sum of the resulting numbers is 0. For each of the two uneliminated polynomials, add the coefficients, reversing the signs of the $x^{3}$ $\displaystyle x^{3}$ and $\displaystyle x$ coefficients:

$x^{4} - 13x ^{3} + 7x ^{2}+ 14x - 9$ $\displaystyle x^{4} - 13x ^{3} + 7x ^{2}+ 14x - 9$:

$\displaystyle 1 + 13 +7 +( - 14) - 9 = -2$

$x^{4} -11x ^{3} + 6x^{2} + 11x - 7$ $\displaystyle x^{4} -11x ^{3} + 6x^{2} + 11x - 7$:

$\displaystyle 1 +11 + 6 +( - 11) (- 7) = 0$

The last polynomial is divisible by both x - 1 $\displaystyle x - 1$ and x+1 $\displaystyle x+1$, and, as a consequence, by $x^{2} - 1$ $\displaystyle x^{2} - 1$.

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Example Question #83 : Single Variable Algebra

Select the polynomial that is divisible by the binomial x - 1 $\displaystyle x - 1$.

Possible Answers:

None of these

$x^{4} + 8x^{3}- 9 x^{2} - 11 x + 10$ $\displaystyle x^{4} + 8x^{3}- 9 x^{2} - 11 x + 10$

$x^{4} + 8x^{3} - 5x^{2} + 9 x -12$ $\displaystyle x^{4} + 8x^{3} - 5x^{2} + 9 x -12$

$x^{4} - 5x^{3} + 8x^{2} - 14x + 10$ $\displaystyle x^{4} - 5x^{3} + 8x^{2} - 14x + 10$

$x^{4} - 8x^{3} + 13x^{2} + 3x - 8$ $\displaystyle x^{4} - 8x^{3} + 13x^{2} + 3x - 8$

Correct answer:

$x^{4} - 5x^{3} + 8x^{2} - 14x + 10$ $\displaystyle x^{4} - 5x^{3} + 8x^{2} - 14x + 10$

Explanation:

A polynomial is divisible by x - 1 $\displaystyle x - 1$ if and only if the sum of its coefficients is 0. Add the coefficients for each given polynomial.

$x^{4} + 8x^{3} - 5x^{2} + 9 x -12$ $\displaystyle x^{4} + 8x^{3} - 5x^{2} + 9 x -12$:

$\displaystyle 1+ 8 +(- 5) + 9+ (- 12) = 1$

$x^{4} + 8x^{3}- 9 x^{2} - 11 x + 10$ $\displaystyle x^{4} + 8x^{3}- 9 x^{2} - 11 x + 10$:

$\displaystyle 1+ 8 + (-9) + (-11)+10 = -1$

$x^{4} - 8x^{3} + 13x^{2} + 3x - 8$ $\displaystyle x^{4} - 8x^{3} + 13x^{2} + 3x - 8$

$\displaystyle 1 + (- 8 )+13 +3+ (- 8 )= 1$

$x^{4} - 5x^{3} + 8x^{2} - 14x + 10$ $\displaystyle x^{4} - 5x^{3} + 8x^{2} - 14x + 10$

$\displaystyle 1+ (- 5) + 8 +(-14) + 10 = 0$

Since its coefficients add up to 0, $x^{4} - 5x^{3} + 8x^{2} - 14x + 10$ $\displaystyle x^{4} - 5x^{3} + 8x^{2} - 14x + 10$ is the only one of the given polynomials divisible by x - 1 $\displaystyle x - 1$.

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Example Question #531 : Sat Subject Test In Math Ii

Which of the following choices gives a sixth root of seven hundred and twenty-nine?

Possible Answers:

$\frac{3 }{2} + \frac{ 3 \sqrt{3 }}{2} i$ $\displaystyle \frac{3 }{2} + \frac{ 3 \sqrt{3 }}{2} i$

$3 + \frac{ 3 \sqrt{3 }}{2} i$ $\displaystyle 3 + \frac{ 3 \sqrt{3 }}{2} i$

None of these

$\frac{3 }{2} + 3 i \sqrt{3 }$ $\displaystyle \frac{3 }{2} + 3 i \sqrt{3 }$

$3 + 3 i \sqrt{3 }$ $\displaystyle 3 + 3 i \sqrt{3 }$

Correct answer:

$\frac{3 }{2} + \frac{ 3 \sqrt{3 }}{2} i$ $\displaystyle \frac{3 }{2} + \frac{ 3 \sqrt{3 }}{2} i$

Explanation:

Let $\displaystyle x$ be a sixth root of 729. The question is to find a solution of the equation

$x^{6} = 729$ $\displaystyle x^{6} = 729$.

Subtracting 64 from both sides, this equation becomes

$x^{6} - 729= 0$ $\displaystyle x^{6} - 729= 0$

729 is a perfect square (of 27) The binomial at left can be factored first as the difference of two squares:

$(x^{3} ) ^{2}- 27 ^{2}= 0$ $\displaystyle (x^{3} ) ^{2}- 27 ^{2}= 0$

$( x^{3} +27 )( x^{3} - 27)= 0$ $\displaystyle ( x^{3} +27 )( x^{3} - 27)= 0$

27 is a perfect cube (of 3), so the two binomials can be factored as the sum and difference, respectively, of two cubes:

$x^{3} + 27 = x^{3}+ 3 ^{3} = (x+3) (x^{2}- x \cdot 3 + 3 ^{2}) = (x+3) (x^{2}-3x+9)$ $\displaystyle x^{3} + 27 = x^{3}+ 3 ^{3} = (x+3) (x^{2}- x \cdot 3 + 3 ^{2}) = (x+3) (x^{2}-3x+9)$

$x^{3} - 27 = x^{3}+ 3 ^{3} = (x-3) (x^{2}+x \cdot 3 + 3 ^{2}) = (x-3) (x^{2}+3x+9)$ $\displaystyle x^{3} - 27 = x^{3}+ 3 ^{3} = (x-3) (x^{2}+x \cdot 3 + 3 ^{2}) = (x-3) (x^{2}+3x+9)$

The equation therefore becomes

$(x+3) (x^{2}-3x+9) (x-3) (x^{2}+3x+9) = 0$ $\displaystyle (x+3) (x^{2}-3x+9) (x-3) (x^{2}+3x+9) = 0$.

By the Zero Product Principle, one of these factors must be equal to 0.

If $\displaystyle x+ 3 = 0$, then x = -3 $\displaystyle x = -3$; if x-3 = 0 $\displaystyle x-3 = 0$, then x = 3 $\displaystyle x = 3$. Therefore, $\displaystyle -3$ and 3 are sixth roots of 729. However, these are not choices, so we examine the other polynomials for their zeroes.

If $x^{2}-3x+9= 0$ $\displaystyle x^{2}-3x+9= 0$, then, setting $\displaystyle a= 1, b=-3, c= 9$ in the following quadratic formula:

$x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ $\displaystyle x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

$= \frac{- (-3) \pm \sqrt{ (-3)^{2}-4(1)(9)}}{2(1)}$ $\displaystyle = \frac{- (-3) \pm \sqrt{ (-3)^{2}-4(1)(9)}}{2(1)}$

$= \frac{3 \pm \sqrt{ 9 -36}}{2 }$ $\displaystyle = \frac{3 \pm \sqrt{ 9 -36}}{2 }$

$= \frac{3 \pm \sqrt{ -27 }}{2 }$ $\displaystyle = \frac{3 \pm \sqrt{ -27 }}{2 }$

$= \frac{3 \pm\sqrt{9 } \cdot \sqrt{ -1} \cdot \sqrt{3 }}{2}$ $\displaystyle = \frac{3 \pm\sqrt{9 } \cdot \sqrt{ -1} \cdot \sqrt{3 }}{2}$

$= \frac{3 \pm 3i \sqrt{3 }}{2}$ $\displaystyle = \frac{3 \pm 3i \sqrt{3 }}{2}$

$= \frac{3 }{2} \pm \frac{ 3 \sqrt{3 }}{2} i$ $\displaystyle = \frac{3 }{2} \pm \frac{ 3 \sqrt{3 }}{2} i$

If $x^{2}+3x+9= 0$ $\displaystyle x^{2}+3x+9= 0$, then, setting $\displaystyle a= 1, b=3, c= 9$ in the quadratic formula:

$x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ $\displaystyle x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

$= \frac{- 3 \pm \sqrt{ 3^{2}-4(1)(9)}}{2(1)}$ $\displaystyle = \frac{- 3 \pm \sqrt{ 3^{2}-4(1)(9)}}{2(1)}$

$= \frac{-3 \pm \sqrt{ 9 -36}}{2 }$ $\displaystyle = \frac{-3 \pm \sqrt{ 9 -36}}{2 }$

$= \frac{-3 \pm \sqrt{ -27 }}{2 }$ $\displaystyle = \frac{-3 \pm \sqrt{ -27 }}{2 }$

$= \frac{-3 \pm\sqrt{9 } \cdot \sqrt{ -1} \cdot \sqrt{3 }}{2}$ $\displaystyle = \frac{-3 \pm\sqrt{9 } \cdot \sqrt{ -1} \cdot \sqrt{3 }}{2}$

$= \frac{-3 \pm 3i \sqrt{3 }}{2}$ $\displaystyle = \frac{-3 \pm 3i \sqrt{3 }}{2}$

$=- \frac{3 }{2} \pm \frac{ 3 \sqrt{3 }}{2} i$ $\displaystyle =- \frac{3 }{2} \pm \frac{ 3 \sqrt{3 }}{2} i$

Therefore, the set of sixth roots of 729 is

$\left \{ -3, 3,- \frac{3 }{2} - \frac{ 3 \sqrt{3 }}{2} i ,- \frac{3 }{2} + \frac{ 3 \sqrt{3 }}{2} i , \frac{3 }{2} - \frac{ 3 \sqrt{3 }}{2} i, \frac{3 }{2} + \frac{ 3 \sqrt{3 }}{2} i \right \}$ $\displaystyle \left \{ -3, 3,- \frac{3 }{2} - \frac{ 3 \sqrt{3 }}{2} i ,- \frac{3 }{2} + \frac{ 3 \sqrt{3 }}{2} i , \frac{3 }{2} - \frac{ 3 \sqrt{3 }}{2} i, \frac{3 }{2} + \frac{ 3 \sqrt{3 }}{2} i \right \}$

Of the choices given, $\frac{3 }{2} + \frac{ 3 \sqrt{3 }}{2} i$ $\displaystyle \frac{3 }{2} + \frac{ 3 \sqrt{3 }}{2} i$ is the one that appears in this set.

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SAT II Math II : SAT Subject Test in Math II

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