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SAT II Math II : Factoring and Finding Roots

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Example Questions

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Example Question #2 :Factoring Polynomials

Factor the trinomial.

$6x^{2}-31x+35$

Possible Answers:

(3x-7)(2x+5)

(2x-7)(3x-5)

(2x+7)(3x-5)

(2x-7)(3x+5)

(3x-7)(2x-5)

Correct answer:

(2x-7)(3x-5)

Explanation:

Use the-method to split the middle term into the sum of two terms whose coefficients have sumand product 6*35 = 210 . These two numbers can be found, using trial and error, to beand.

-21*-10=210 and -21+(-10)=-31

Now we know that $6x^{2}-31x+35$ is equal to $6x^{2}-10x-21x+35$ .

Factor by grouping.

$(6x^{2}-10x)-(21x-35)$

2x(3x-5)-7(3x-5)

(2x-7)(3x-5)

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Example Question #1 :Factoring And Finding Roots

Factor completely:

$t^{3} + 343$

Possible Answers:

$(t + 49)(t^{2}-7t+ 7)$

$(t + 7)(t-7)^{2}$

The polynomial is prime.

$(t + 7)(t^{2}-7t+ 49)$

$\left ( t+7\right )^{3}$

Correct answer:

$(t + 7)(t^{2}-7t+ 49)$

Explanation:

Since both terms are perfect cubes $\left (343 = 7^{3} \right )$ , the factoring pattern we are looking to take advantage of is the sum of cubes pattern. This pattern is

$A^{3}+ B^{3} = (A + B) (A^{2}-AB +B^{2})$

We substituteforand 7 for:

$t^{3} + 343 = t^{3} + 7^{3}$

$= (t+ 7) (t^{2}-t \cdot 7+7^{2})$

$= (t+ 7) (t^{2}-7t+49)$

The latter factor cannot be factored further, since we would need to find two integers whose product is 49 and whose sum is; they do not exist. This is as far as we can go with the factoring.

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Example Question #1 :Factoring And Finding Roots

Which of the following values ofwould make

$t^{6}+N$

a prime polynomial?

Possible Answers:

None of the other responses is correct.

125

Correct answer:

None of the other responses is correct.

Explanation:

$t^{6}$ is the cube of $t^{2}$ . Therefore, ifis a perfect cube, the expression $t^{6}+N$ is factorable as the sum of two cubes. All four of the choices are perfect cubes - 8, 27, 64, and 125 are the cubes of 2, 3, 4 and 5, respectively. The correct response is that none of the choices are correct.

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Example Question #3 :Factoring And Finding Roots

Which of the following values ofwouldnotmake

$9t^{2}+N$

a prime polynomial?

Possible Answers:

None of the other responses is correct.

N = 64

N = 25

N = 36

N = 49

Correct answer:

N = 36

Explanation:

$9t^{2}$ is a perfect square term - it is equal to $\left (3t \right )^{2}$ . All of the values ofgiven in the choices are perfect squares - 25, 36, 49, and 64 are the squares of 5, 6, 7, and 8, respectively.

Therefore, for each given value of, the polynomial is the sum of squares, which is normally a prime polynomial. However, if N = 36 - and only in this case - the polynomial can be factored as follows:

$9t^{2}+36 = 9 \cdot t^{2} + 9 \cdot 4 = 9 (t^{2}+ 4)$ .

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Example Question #4 :Factoring And Finding Roots

Which of the following is a factor of the polynomial $x^{4} - 11x^{2} + 23x - 18$ ?

Possible Answers:

x - 3

x-4

x- 1

x- 2

x- 5

Correct answer:

x- 2

Explanation:

Call $P(x) = x^{4} - 11x^{2} + 23x - 18$ .

By the Rational Zeroes Theorem, since P (x) has only integer coefficients, any rational solution of P (x) = 0 must be a factor of 18 divided by a factor of 1 - positive or negative. 18 has as its factors 1, 2, 3, 6, 9, and 18; 1 has only itself as a factor. Therefore, the rational solutions ofmust be chosen from this set:

$\left \{ 1, 2, 3, 6, 9, 18 -1,- 2,- 3,- 6, -9, -18 \right \}$ .

By the Factor Theorem, a polynomial P (x) is divisible by x- c if and only if P(c) = 0 - that is, ifis a zero. By the preceding result, we can immediately eliminate x-4 and x- 5 as factors, since 4 and 5 have been eliminated as possible zeroes.

Of the three remaining choices, we can demonstrate that x- 2 is the factor by evaluating P(2) :

$P(x) = x^{4} - 11x^{2} + 23x - 18$

$P(2) = 2^{4} - 11 \cdot 2 ^{2} + 23 \cdot 2 - 18$

$= 16 - 11 \cdot 4 + 23 \cdot 2 - 18$

= 16 - 44 + 46 - 18

= 0

By the Factor Theorem, it follows that x- 2 is a factor.

As for the other two, we can confirm that neither is a factor by evaluating P(1) and P(3) :

$P(1) = 1^{4} - 11 \cdot 1 ^{2} + 23 \cdot 1 - 18$

$= 1- 11 \cdot 1 + 23 \cdot 1 - 18$

= 1- 11 + 23 - 18

= -5

$P(3) = 3^{4} - 11 \cdot 3 ^{2} + 23 \cdot 3 - 18$

$= 81 -11\cdot 9 + 23 \cdot 3 - 18$

= 81 -99 + 69 - 18

=33

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Example Question #2 :Factoring And Finding Roots

Give the set of all real solutions of the equation $2x^{4} - 13x^{2} +21 = 0$ .

Possible Answers:

$\left \{ - \frac{\sqrt{14}}{2}, \frac{ \sqrt{14}}{2} \right \}$

$\left \{ \sqrt{3} ,\frac{ \sqrt{14}}{2} \right \}$

$\left \{ - \sqrt{3}, \sqrt{3} \right \}$

The equation has no real solution.

$\left \{ - \frac{\sqrt{14}}{2}, - \sqrt{3}, \sqrt{3} ,\frac{ \sqrt{14}}{2} \right \}$

Correct answer:

$\left \{ - \frac{\sqrt{14}}{2}, - \sqrt{3}, \sqrt{3} ,\frac{ \sqrt{14}}{2} \right \}$

Explanation:

Set $y = x^{2}$ . Then $x^{4 } = \left (x^{2} \right )^{2} = y ^{2}$ .

$2x^{4} - 13x^{2} +21 = 0$ can be rewritten as

$2 \left (x^{2} \right )^{2} - 13x^{2} +21 = 0$

Substituting $y^{2}$ for $x^{4 }$ andfor $x^{2}$ , the equation becomes

$2 y^{2} - 13y +21 = 0$ ,

a quadratic equation in.

This can be solved using themethod. We are looking for two integers whose sum isand whose product is 2 (21)= 42 . Through some trial and error, the integers are found to beand, so the above equation can be rewritten, and solved using grouping, as

$2 y^{2} - 7y - 6y +21 = 0$

$(2 y^{2} - 7y )- (6y -21 )= 0$

y(2 y - 7 )- 3(2 y - 7 ) = 0

(y - 3 )(2 y - 7 ) = 0

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

Either:

y- 3 = 0

y - 3 + 3 = 0 + 3

y = 3

Substituting $x^{2}$ back for:

$x^{2} = 3$

Taking the positive and negative square roots of both sides:

$x = \pm \sqrt{3}$ .

Or:

2y- 7 = 0

2y- 7+ 7 = 0 + 7

2y = 7

$\frac{2y}{2} =\frac{ 7}{2}$

$y =\frac{ 7}{2}$

Substituting back:

$x^{2} =\frac{ 7}{2}$

Taking the positive and negative square roots of both sides, and applying the Quotient of Radicals property, then simplifying by rationalizing the denominator:

$x =\pm \sqrt{\frac{ 7}{2}} = \pm \frac{\sqrt{7}}{\sqrt{2}} = \pm \frac{\sqrt{7} \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2} } =\pm \frac{\sqrt{14}}{2}$

The solution set is $\left \{ - \frac{\sqrt{14}}{2}, - \sqrt{3}, \sqrt{3} ,\frac{ \sqrt{14}}{2} \right \}$ .

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Example Question #6 :Factoring And Finding Roots

Define a function $f(x) = x^{2} + \cos 3x$ .

f(c) = 4 for exactly one real value ofon the interval (1.5, 2.0) .

Which of the following statements is correct about?

Possible Answers:

$c \in (1.9, 2.0)$

$c \in (1.6, 1.7)$

$c \in (1.8, 1.9)$

$c \in (1.7, 1.8)$

$c \in (1.5, 1.6)$

Correct answer:

$c \in (1.8, 1.9)$

Explanation:

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

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

Evaluate g(x) for each of the following values: $\left \{ 1.5, 1.6, 1.7,1.8, 1.9, 2.0 \right \}$

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

$g(1.5)= 1.5^{2} + \cos 3 (1.5) - 4$

$= 2.25 + \cos 4.5 - 4$

$\approx 2.25 + (-0.21 )- 4$

$\approx - 1.96$

$g(1.6)= 1.6^{2} + \cos 3 (1.6) - 4$

$= 2.56 + \cos 4.8 - 4$

$\approx 2.56 +0.09 - 4$

$\approx - 1.35$

$g(1.7)= 1.7^{2} + \cos 3 (1.7) - 4$

$=2.89 + \cos 5.1- 4$

$\approx 2.89 + 0.38- 4$

$\approx - 0.73$

$g(1.8)= 1.8^{2} + \cos 3 (1.8) - 4$

$= 3.24 + \cos 5.4 - 4$

$\approx 3.24 + 0.63 - 4$

$\approx - 0.13$

$g(1.9)= 1.9^{2} + \cos 3 (1.9) - 4$

$=3.61 + \cos 5.7 - 4$

$\approx 3.61 + 0.83 - 4$

$\approx 0.44$

$g(2.0)= 2.0^{2} + \cos 3 (2.0) - 4$

$= 4 + \cos 6 - 4$

$\approx 4 +0.96 - 4$

$\approx 0.96$

Only in the case of (1.8, 1.9) does it hold that g (x) assumes a different sign at each endpoint - g(1.8) < 0 < g(1.9) . By the IVT, g(c) = 0 , and f(c) = 4 , for some $c \in (1.8, 1.9)$ .

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Example Question #7 :Factoring And Finding Roots

A cubic polynomial p(x) with rational coefficients whose lead term is $x^{3}$ has 2 + 3i and 2 - 3i as two of its zeroes. Which of the following is this polynomial?

Possible Answers:

The correct answer cannot be determined from the information given.

$p(x)= x^{3} -2x^{2}+5x+26$

$p(x)= x^{3} -3x^{2}+9x+13$

$p(x)= x^{3} -5x^{2}+17x-13$

$p(x)= x^{3} -6x^{2}+21x-26$

Correct answer:

The correct answer cannot be determined from the information given.

Explanation:

与有理系数多项式s imaginary zeroes in conjugate pairs. Two imaginary zeroes are given that are each other's complex conjugate - 2 + 3i and 2 -3i . Since the polynomial is cubic - of degree 3 - it has one other zero, which must be real. However, no information is given about that zero. Therefore, the polynomial cannot be determined.

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Example Question #8 :Factoring And Finding Roots

Define functions p (x) =7x and $q (x) = 5^{x}$ .

p(c) =q(c) for exactly one value ofon the interval (1, 1.5) . Which of the following is true of?

Possible Answers:

$c\ (1.1, 1.2)$

$c \in (1.0, 1.1)$

$c \in (1.4, 1.5)$

$c \in (1.2, 1.3 )$

$c \in (1.3, 1.4)$

Correct answer:

$c \in (1.4, 1.5)$

Explanation:

Define

g(x) = p(x) - q(x)

$= 7x- 5^{x}$

Then if g(c) = 0 ,

it follows that

p(c) - q(c) = 0 ,

or, equivalently,

p(c) = q(c) .

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

Since polynomial p(x) and exponential function g(x) are continuous everywhere, so is g(x) , so the IVT applies here.

Evaluate g(x) for each of the following values: $\left \{ 1. 0 , 1.1, 1.2, 1.3, 1.4, 1.5 \right \}$ :

$g(x) = 7x- 5^{x}$

$g(1) = 7 (1)- 5^{1} = 7 - 5 = 2$

$g(1.1) = 7 (1.1)- 5^{1.1} \approx 7.7 - 5.9 \approx 1.8$

$g(1.2) = 7 (1.2)- 5^{1.2} \approx 8.4 - 6.9 \approx 1.5$

$g(1.3) = 7 (1.3)- 5^{1.3} \approx 9.1 - 8.1 \approx 1$

$g(1.4) = 7 (1.4)- 5^{1.4} \approx 9.8 - 9.5 \approx 0.3$

$g(1.5) = 7 (1.5)- 5^{1.5} \approx 10.5- 11.2 \approx - 0.7$

Only in the case of (1.4, 1.5) does it hold that g (x) assumes a different sign at both endpoints - g(1.4)> 0 > g(1.5) . By the IVT, g(c) = 0 , and p(c) = q(c) , for some $c \in (1.4, 1.5)$ .

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Example Question #9 :Factoring And Finding Roots

A cubic polynomial p(x) with rational coefficients and with $x^{3}$ as its leading term has 2 and 3 as its only zeroes. 2 is a zero of multiplicity 1.

Which of the following is this polynomial?

Possible Answers:

$p(x) = x^{3} - 7x^{2} + 16x- 12$

$p(x) = x^{3} + 7x^{2} + 16x + 12$

Insufficient information exists to determine the polynomial.

$p(x) = x^{3} - 8x^{2} + 21x - 18$

$p(x) = x^{3} + 8x^{2} + 21x + 18$

Correct answer:

$p(x) = x^{3} - 8x^{2} + 21x - 18$

Explanation:

A cubic polynomial has three zeroes, if a zero of multiplicityis countedtimes. Since its lead term is $x^{3}$ , we know that, in factored form,

$p(x) = (x-b_{1}) (x-b_{2}) (x-b_{3})$ ,

where $b_{1}$ , $b_{2}$ , and $b_{3}$ are its zeroes.

Since 2 is a zero of multiplicity 1, its only other zero, 3, must be a zero of multiplicity 2.

Therefore, we can set $b_{1} = 2$ , $b_{2} = b_{3}= 3$ , in the factored form of p(x) , and

p(x) = (x-2) (x-3) (x-3) ,

$p(x) = (x-2) (x-3) ^{2}$

To rewrite this, firs square x -3 by way of the square of a binomial pattern:

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

$(x-3)^{2} = x^{2} - 6x + 9$

Thus,

$p(x) = (x-2) ( x^{2} - 6x + 9)$

Multiplying:

$x^{2} - 6x + 9$

$\underline{x - 2}$

$-2x^{2} +12x - 18$

$\underline{x^{3} - 6x^{2} + 9x}$ ________

$x^{3} - 8x^{2} + 21x - 18$ ,

the correct polynomial.

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SAT II Math II : Factoring and Finding Roots

Study concepts, example questions & explanations for SAT II Math II

All SAT II Math II Resources

Example Questions

Example Question #2 :Factoring Polynomials

Example Question #1 :Factoring And Finding Roots

Example Question #1 :Factoring And Finding Roots

Example Question #3 :Factoring And Finding Roots

Example Question #4 :Factoring And Finding Roots

Example Question #2 :Factoring And Finding Roots

Example Question #6 :Factoring And Finding Roots

Example Question #7 :Factoring And Finding Roots

Example Question #8 :Factoring And Finding Roots

Example Question #9 :Factoring And Finding Roots

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