Factoring and Finding Roots - SAT Subject Test in Math II

Card 0 of 20

Question

Factor the trinomial.

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

Answer

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

and

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

Factor by grouping.

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

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Question

Factor completely:

$t^{3} + 343$

Answer

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 substitute for and 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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Question

Which of the following values of would make

$t^{6}+N$

a prime polynomial?

Answer

$t^{6}$ is the cube of $t^{2}$ . Therefore, if is 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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Question

Which of the following values of would not make

$9t^{2}+N$

a prime polynomial?

Answer

$9t^{2}$ is a perfect square term - it is equal to $\left (3t \right )^{2}$ . All of the values of given 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 - 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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Question

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

Answer

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 $x^{4 }$ and for $x^{2}$ , the equation becomes

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

a quadratic equation in .

This can be solved using the method. We are looking for two integers whose sum is and whose product is . Through some trial and error, the integers are found to be and , so the above equation can be rewritten, and solved using grouping, as

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

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

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

Either:

Substituting $x^{2}$ back for :

$x^{2} = 3$

Taking the positive and negative square roots of both sides:

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

Or:

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

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

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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Question

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

Answer

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

By the Rational Zeroes Theorem, since has only integer coefficients, any rational solution of 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 of must 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 is divisible by if and only if - that is, if is a zero. By the preceding result, we can immediately eliminate and as factors, since 4 and 5 have been eliminated as possible zeroes.

Of the three remaining choices, we can demonstrate that is the factor by evaluating :

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

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

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

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

As for the other two, we can confirm that neither is a factor by evaluating and :

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

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

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

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

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Question

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

for exactly one real value of on the interval .

Which of the following statements is correct about ?

Answer

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

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

Evaluate 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 does it hold that assumes a different sign at each endpoint - . By the IVT, , and , for some $c \in (1.8, 1.9)$ .

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Question

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

Answer

A polynomial with rational coefficients has its imaginary zeroes in conjugate pairs. Two imaginary zeroes are given that are each other's complex conjugate - and . 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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Question

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

for exactly one value of on the interval . Which of the following is true of ?

Answer

Define

$= 7x- 5^{x}$

Then if ,

it follows that

,

or, equivalently,

.

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

Since polynomial and exponential function are continuous everywhere, so is , so the IVT applies here.

Evaluate 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 does it hold that assumes a different sign at both endpoints - . By the IVT, , and , for some $c \in (1.4, 1.5)$ .

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Question

A cubic polynomial 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?

Answer

A cubic polynomial has three zeroes, if a zero of multiplicity is counted times. 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 , and

,

or

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

To rewrite this, firs square 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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Question

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

Answer

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

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

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

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

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

a quadratic equation in .

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

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

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

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

Either:

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

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

Substituting back:

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

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

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

Or:

Substituting back:

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

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

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Question

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

for exactly one positive value of ; this is on the interval . Which of the following is true of ?

Answer

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

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

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

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

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

$=0 - \cos 0- 3$

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

$=1 - \cos 4- 3$

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

$\approx -1.35$

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

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

$\approx 1.15$

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

$\approx 9- 0.84 - 3$

$\approx 5.16$

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

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

$\approx 13.96$

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

$\approx 25 - 0.41 - 3$

$\approx 21.59$

Only in the case of does it hold that assumes a different sign at the endpoints - . By the IVT, , and , for some $c \in (1, 2)$ .

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Question

Which of the following is a cube root of ?

Answer

Let be a cube root of . The question is to find a solution of the equation

$x^{3} =- 64$ .

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

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

$x^{3} + 64 =0$

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

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

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

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

We can set both factors equal to zero and solve:

is a cube root of ; however, this is not one of the choices.

Setting

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

we can make use of the quadratic formula, setting in the following:

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

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

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

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

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

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

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

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

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

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Question

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

Answer

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

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

A polynomial with rational coefficients has its imaginary zeroes in conjugate pairs. Since is such a polynomial, then, since is a zero, so is its complex conjugate ; similarly, since is a zero, so is its complex conjugate . Substituting these four values for the four $b_{n}$ values:

This can be rewritten as

or

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

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

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

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

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

Multiply the last two factors similarly:

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

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

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

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

Thus,

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

Multiply:

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

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

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

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

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

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

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Question

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

for exactly one value of on the interval . Which statement is true about ?

Answer

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

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

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

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

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

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

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

$=64 - 7 \cdot 16 -5$

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

$=729 - 7 \cdot 81 -5$

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

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

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

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

Only in the case of does it hold that assumes different signs at the endpoints - . By the IVT, , and , for some $c \in (2,3)$ .

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Question

Select the polynomial that is divisible by the binomial .

Answer

A polynomial is divisible by if and only if the sum of its coefficients is 0. Add the coefficients for each given polynomial.

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

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

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

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

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

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Question

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

Answer

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

$x^{6} = 729$ .

Subtracting 64 from both sides, this equation becomes

$x^{6} - 729= 0$

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

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

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

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

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

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

The equation therefore becomes

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

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

If , then ; if , then . Therefore, and 3 are sixth roots of 729. However, these are not choices, so we examine the other polynomials for their zeroes.

If $x^{2}-3x+9= 0$ , then, setting in the following quadratic formula:

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

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

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

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

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

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

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

If $x^{2}+3x+9= 0$ , then, setting in the quadratic formula:

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

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

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

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

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

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

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

Therefore, the set of sixth roots of 729 is

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

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

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Question

A polynomial of degree 4 has as its lead term $x^{4}$ and has rational coefficients. One of its zeroes is ; this zero has multiplicity two.

Which of the following is this polynomial?

Answer

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

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

where the $b_{n}$ terms are the four zeroes.

A polynomial with rational coefficients has its imaginary zeroes in conjugate pairs. Since is such a polynomial, then, since is a zero of multiplicity 2, so is its complex conjugate . We can set $b_{1} = b_{3} = 4-i$ and $b_{2} = b_{4} = 4+i$ , and

$p(x) =\left {[x-(3+ 2i)] [x-(3- 2i)] \right }^{2}$

We can rewrite this as

$p(x) = \left { [x- 3- 2i][x- 3+ 2i ] \right } ^{2}$

or

$p(x) =\left { [(x- 3)-2i ][(x- 3)+2i ] \right } ^{2}$

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

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

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

$= x ^{2} -6x+9 -(4 i^{2})$

$= x ^{2} -6x+9 -(-4 )$

$= x ^{2} -6x+13$

Therefore,

$p(x) =( x ^{2} -6x+13 )^{2}$

Multiplying:

$x ^{2} -6x+13$

$\underline{x ^{2} -6x+13}$

$13 x^{2} -78x +169$

$-6x^{3} +36 x ^{2} -78x$

$\underline{x^{4} -6x^{3} +13x^{2}\textup{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }}$

$x^{4} - 12x^{3} + 62x^{2} - 156x +169$

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Question

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

Answer

One way to work this problem is as follows:

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

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

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

A polynomial is divisible by if and only if the sum of its coefficients is 0. Add the coefficients for each given polynomial:

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

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

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

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

The last two polynomials are both divisible by . The other two can be eliminated as correct choices.

A polynomial is divisible by 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 $x^{3}$ and coefficients:

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

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

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

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Question

What is a possible root to ?

Answer

Factor the trinomial.

The multiples of the first term is .

The multiples of the third term is .

We can then factor using these terms.

Set the equation to zero.

This means that each product will equal zero.

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

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

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