All SAT II Math II Resources
Example Questions
Example Question #1 : How To Factor A Trinomial
Factor the trinomial.
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 is equal to .
Factor by grouping.
Example Question #1 : Factoring And Finding Roots
Factor completely:
The polynomial is prime.
Since both terms are perfect cubes , the factoring pattern we are looking to take advantage of is the sum of cubes pattern. This pattern is
We substitute for and 7 for :
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.
Example Question #3 : Factoring And Finding Roots
Which of the following values of would make
a prime polynomial?
None of the other responses is correct.
None of the other responses is correct.
is the cube of . Therefore, if is a perfect cube, the expression 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.
Example Question #4 : Factoring And Finding Roots
Which of the following values of would not make
a prime polynomial?
None of the other responses is correct.
is a perfect square term - it is equal to . 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:
.
Example Question #5 : Factoring And Finding Roots
Which of the following is a factor of the polynomial ?
Call .
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:
.
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 :
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 :
Example Question #6 : Factoring And Finding Roots
Give the set of all real solutions of the equation .
The equation has no real solution.
Set . Then .
can be rewritten as
Substituting for and for , the equation becomes
,
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
By the Zero Product Principle, one of these factors is equal to zero:
Either:
Substituting back for :
Taking the positive and negative square roots of both sides:
.
Or:
Substituting back:
Taking the positive and negative square roots of both sides, and applying the Quotient of Radicals property, then simplifying by rationalizing the denominator:
The solution set is .
Example Question #7 : Factoring And Finding Roots
Define a function .
for exactly one real value of on the interval .
Which of the following statements is correct about ?
Define . 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 . is a continuous function, so the IVT applies here.
Evaluate for each of the following values:
Only in the case of does it hold that assumes a different sign at each endpoint - . By the IVT, , and , for some .
Example Question #6 : Factoring And Finding Roots
A cubic polynomial with rational coefficients whose lead term is has and as two of its zeroes. Which of the following is this polynomial?
The correct answer cannot be determined from the information given.
The correct answer cannot be determined from the information given.
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.
Example Question #8 : Factoring And Finding Roots
Define functions and .
for exactly one value of on the interval . Which of the following is true of ?
Define
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 .
Since polynomial and exponential function are continuous everywhere, so is , so the IVT applies here.
Evaluate for each of the following values: :
Only in the case of does it hold that assumes a different sign at both endpoints - . By the IVT, , and , for some .
Example Question #3 : Factoring And Finding Roots
A cubic polynomial with rational coefficients and with 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?
Insufficient information exists to determine the polynomial.
A cubic polynomial has three zeroes, if a zero of multiplicity is counted times. Since its lead term is , we know that, in factored form,
,
where , , and 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 , , in the factored form of , and
,
or
To rewrite this, firs square by way of the square of a binomial pattern:
Thus,
Multiplying:
________
,
the correct polynomial.
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