SAT Math - Factoring and Finding Roots

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:

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

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 $c \in (1.4, 1.5)$ .

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 $c \in (1, 2)$ .

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:

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

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

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

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

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

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

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

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

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

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

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

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

None of these

Explanation

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 $x^{2} - 1$ .

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?

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

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

Insufficient information exists to determine the polynomial.

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

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

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 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: