DSQ: Understanding factoring - GMAT Quantitative Reasoning
Card 0 of 9
If 
and 
are positive, what is the value of 
?
(1) 
(2) 
If
(1)
(2)
For statement (1), we can factor the equation as following: 
. Obviously we cannot figure out 
since we have no information about the value of 
.
From statement (2) only, we have no idea what 
is by knowing the value of 
. However, putting the two statements together, we will get 
.
For statement (1), we can factor the equation as following:
From statement (2) only, we have no idea what
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Willy's teacher challenged him to write two whole numbers in the square and circle in the diagram below in order to make a polynomial that could be factored.
Did Willy succeed?
Statement 1: The number Willy wrote in the square is a multple of 4.
Statement 2: The number Willy wrote in the circle is a multiple of 9.
Willy's teacher challenged him to write two whole numbers in the square and circle in the diagram below in order to make a polynomial that could be factored.
Did Willy succeed?
Statement 1: The number Willy wrote in the square is a multple of 4.
Statement 2: The number Willy wrote in the circle is a multiple of 9.
Assume both statements are true.
If Willy wrote a 4 in the circle and a 9 in the square, both statements are satisfied, and the resulting polynomial can be factored as the difference of squares:
If Willy wrote a 4 in the circle and a 27 in the square, both statements are satisfied. However, the resulting polynomial
is prime; no greatest common factor can be taken out, and the only possible pattern, the difference of squares, does not fit, since the square root of 27 is irrational.
The two statements together are insufficient.
Assume both statements are true.
If Willy wrote a 4 in the circle and a 9 in the square, both statements are satisfied, and the resulting polynomial can be factored as the difference of squares:
If Willy wrote a 4 in the circle and a 27 in the square, both statements are satisfied. However, the resulting polynomial
is prime; no greatest common factor can be taken out, and the only possible pattern, the difference of squares, does not fit, since the square root of 27 is irrational.
The two statements together are insufficient.
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Chad's teacher challenged him to write two whole numbers in the square and circle in the diagram below in order to make a polynomial that could be factored.
Did Chad succeed?
Statement 1: The cube root of the number Chad wrote in the square is a whole number.
Statement 2: The cube root of the number Chad wrote in the circle is an irrational number.
Chad's teacher challenged him to write two whole numbers in the square and circle in the diagram below in order to make a polynomial that could be factored.
Did Chad succeed?
Statement 1: The cube root of the number Chad wrote in the square is a whole number.
Statement 2: The cube root of the number Chad wrote in the circle is an irrational number.
Assume both statements are true. To try to factor the polynomial, we have one of two possibilities - the difference of cubes, or a greatest common factor. But since the number in the circle is not a perfect cube, this leaves us with factoring by GCF.
The polynomial 
fits both statements; 
, so it cannot be factored.
The polynomial 
fits both statements; 
, so it can be factored:
Assume both statements are true. To try to factor the polynomial, we have one of two possibilities - the difference of cubes, or a greatest common factor. But since the number in the circle is not a perfect cube, this leaves us with factoring by GCF.
The polynomial
The polynomial
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Julia's teacher challenged her to write two whole numbers in the square and circle in the diagram below in order to make a polynomial that could be factored.
Did Julia succeed?
Statement 1: The cube root of the number Julia wrote in the circle is a whole number.
Statement 2: The number Julia wrote in the square is ten times the number Julia wrote in the circle.
Julia's teacher challenged her to write two whole numbers in the square and circle in the diagram below in order to make a polynomial that could be factored.
Did Julia succeed?
Statement 1: The cube root of the number Julia wrote in the circle is a whole number.
Statement 2: The number Julia wrote in the square is ten times the number Julia wrote in the circle.
A binomial of the form 
can be factored out as the sum of cubes if and only if both 
and 
are perfect cubes, and it can be factored out by taking out a GCF if the GCF of 
and 
is not 1.
Assume both statements to be true.
- which is equal to 
- fits the conditions. But the polynomial cannot be factored using the sum of cubes property (10 is not a cube, but 1 is), nor can a greatest common factor be taken out (the greatest common factor of the terms is 1).
fits the conditions, and can be factored out as
A binomial of the form
Assume both statements to be true.
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Karen's teacher challenged her to write two whole numbers in the square and circle in the diagram below in order to make a polynomial that could be factored.
Assuming both numbers are whole numbers, did Karen succeed?
Statement 1: The cube root of the number Julia wrote in the circle is a whole number.
Statement 2: The number Julia wrote in the square is twenty-seven times the number Julia wrote in the circle.
Karen's teacher challenged her to write two whole numbers in the square and circle in the diagram below in order to make a polynomial that could be factored.
Assuming both numbers are whole numbers, did Karen succeed?
Statement 1: The cube root of the number Julia wrote in the circle is a whole number.
Statement 2: The number Julia wrote in the square is twenty-seven times the number Julia wrote in the circle.
A binomial of the form can be factored out as the sum of cubes if and only if both and are perfect cubes, and it can be factored out by taking out a GCF if the GCF of and is not 1.
Statement 1 alone gives no clue as to the number that was placed in the square. The polynomials 
and 
fit the statement, but neither can be factored as the sum of cubes, and only the latter can have a greatest common factor (4) factored out.
Assume Statement 2 alone. There are two possible cases:
Case 1: Julia wrote a 1 in the circle.
The polynomial is 
, which, as the sum of cubes, is factorable.
Case 2: Julia wrote a different whole number in the circle.
Since the number in the square is 27 times the number in the circle, at the very least, the polynomial can be factored by distributing out the number in the circle. For example,
A binomial of the form
Statement 1 alone gives no clue as to the number that was placed in the square. The polynomials
Assume Statement 2 alone. There are two possible cases:
Case 1: Julia wrote a 1 in the circle.
The polynomial is
Case 2: Julia wrote a different whole number in the circle.
Since the number in the square is 27 times the number in the circle, at the very least, the polynomial can be factored by distributing out the number in the circle. For example,
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Don's teacher challenged him to write two whole numbers in the square and circle in the diagram below in order to make a polynomial that could be factored.
Assuming that Don wrote two whole numbers, did he succeed?
Statement 1: The number Don wrote in the circle is the square of half the number he wrote in the square.
Statement 2: The sum of the numbers Don wrote in the square and in the circle is 15.
Don's teacher challenged him to write two whole numbers in the square and circle in the diagram below in order to make a polynomial that could be factored.
Assuming that Don wrote two whole numbers, did he succeed?
Statement 1: The number Don wrote in the circle is the square of half the number he wrote in the square.
Statement 2: The sum of the numbers Don wrote in the square and in the circle is 15.
By Statement 1 alone, if we call the number in the square 
, then the number in the circle is the square of half this, or 
; the polynomial is therefore
,
which is the pattern of a perfect square trinomial. The polynomial can be factored, and Don succeeded.
Assume Statement 2 alone. The polynomials
and
fit the conditions of Statement 2.
Some trinomial of the pattern 
can be factored as 
, where 
and 
have product 
and sum 
.
, since 3 and 3 have sum 6 and product 9.
However,
cannot be factored; no two integers exist with product 10 and sum 5.
By Statement 1 alone, if we call the number in the square
which is the pattern of a perfect square trinomial. The polynomial can be factored, and Don succeeded.
Assume Statement 2 alone. The polynomials
and
fit the conditions of Statement 2.
Some trinomial of the pattern
However,
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Theresa's teacher challenged her to write whole numbers in the circle and the square in the diagram below in order to make a polynomial that could be factored.
Assuming Theresa wrote two whole numbers, did she succeed?
Statement 1: Theresa wrote a 64 in the circle.
Statement 2: Theresa wrote a multiple of 6 in the square.
Theresa's teacher challenged her to write whole numbers in the circle and the square in the diagram below in order to make a polynomial that could be factored.
Assuming Theresa wrote two whole numbers, did she succeed?
Statement 1: Theresa wrote a 64 in the circle.
Statement 2: Theresa wrote a multiple of 6 in the square.
The template fits the difference of squares pattern, which may be used only if all coefficients are perfect squares and all exponents are even (making the powers of the variables perfect squares). Each statement only answers the question of one of these qualifications; the two together answer both.
The template fits the difference of squares pattern, which may be used only if all coefficients are perfect squares and all exponents are even (making the powers of the variables perfect squares). Each statement only answers the question of one of these qualifications; the two together answer both.
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Consider function 
.
I) 
has zeroes at 
and 
.
II) 
is a second degree polynomial.
Find the equation that models 
.
Consider function
I)
II)
Find the equation that models
If we consider statement II, we know f(x) must be of the form
.
Due to the zero product property and statement I, we know that f(x) CAN look like this
.
However, it could also look like this
.
Or this
.
So with what we are given it is impossible to find the true equation for f(x).
If we consider statement II, we know f(x) must be of the form
Due to the zero product property and statement I, we know that f(x) CAN look like this
However, it could also look like this
Or this
So with what we are given it is impossible to find the true equation for f(x).
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Simplify: 
Simplify:
In order to simplify, we must first pull out the largest common factor of each term in the numerator, -3:
We then recognize that the denominator is a difference of squares:
We can therefore cancel the (x-1) terms and are left with:
In order to simplify, we must first pull out the largest common factor of each term in the numerator, -3:
We then recognize that the denominator is a difference of squares:
We can therefore cancel the (x-1) terms and are left with:
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