Factoring and Finding Roots

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SAT Math › Factoring and Finding Roots

Questions 1 - 10

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

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

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

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

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

The equation has no real solution.

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 $x^{4 }$ 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 $\left \{ - \frac{\sqrt{14}}{2}, - \sqrt{3}, \sqrt{3} ,\frac{ \sqrt{14}}{2} \right \}$ .

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?

The correct answer cannot be determined from the information given.

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

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

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

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

Explanation

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.

Which of the following values of would not make

$9t^{2}+N$

a prime polynomial?

None of the other responses is correct.

Explanation

$9t^{2}$ 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:

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

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?

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

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

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

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

Insufficient information exists to determine the polynomial.

Explanation

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

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

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.

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

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

None of these

$3 + 3 i \sqrt{3 }$

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

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

Explanation

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.

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?

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

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

$x^{4} +12x^{3} + 46x^{2}+ 60x +25$

$x^{4} -12x^{3} + 46x^{2}- 60x +25$

Cannot be determined

Explanation

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

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

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

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:

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

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

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

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

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

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

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 ?

$c \in (1, 2)$

$c \in (0, 1)$

$c \in (2, 3)$

$c \in (3, 4)$

$c \in (4,5 )$

Explanation

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 $c \in (1, 2)$ .

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

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

$c \in (1.4, 1.5)$

$c \in (1.3, 1.4)$

$c \in (1.2, 1.3 )$

$c\ (1.1, 1.2)$

$c \in (1.0, 1.1)$

Explanation

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 $c \in (1.4, 1.5)$ .

What is a possible root to ?

$-\frac{1}{2}$

$-\frac{3}{2}$

$\frac{2}{9}$

$\frac{1}{2}$

Explanation

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

Find the roots to:

$\pm \frac{\sqrt3}{3}$

$\pm \frac{1}{3}$

$\pm \frac{1}{9}$

$\pm \sqrt{3}$

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

Explanation

This equation cannot be factored. Since this is a parabola, we can use the quadratic equation to find the roots.

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

Substitute the coefficients of .

$x = \frac{-0\pm \sqrt{0^2-4(3)(-1)}}{2(3)} =\pm \frac{\sqrt{12}}{6} = \pm\frac{2\sqrt3}{6}$

The answers are: $\pm \frac{\sqrt3}{3}$

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