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Test: Polynomial Identities - SAT MCQ


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10 Questions MCQ Test Mathematics for SAT - Test: Polynomial Identities

Test: Polynomial Identities for SAT 2024 is part of Mathematics for SAT preparation. The Test: Polynomial Identities questions and answers have been prepared according to the SAT exam syllabus.The Test: Polynomial Identities MCQs are made for SAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Polynomial Identities below.
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Test: Polynomial Identities - Question 1

What do we get after factorising x+ 8y3+ z3 – 4xyz?

Detailed Solution for Test: Polynomial Identities - Question 1

We know that a3 + b3 + c3 – 3abc = (a + b - c)(a2 + b2 + c2 – ab – bc – ca)
In x3 + 8y3 + y3 – 6xyz, a = x, b = 2y and c = z
By using the above equation, we get x3 + 8y3 + z3 – 4xyz = (x + 2y + z) (x2 + (2y)2 + z2 – x(2y) – (2y)(z) – zx)
= (x + 2y + z) (x2 + 4y2 + z2 – 2xy – 2yz – zx).

Test: Polynomial Identities - Question 2

953 = __________ (calculate without direct calculation).

Detailed Solution for Test: Polynomial Identities - Question 2

We know that (x - y)3 = x– y3 – 3 xy (x - y).
953 can also be written as (100-5)3
Therefore, 95= (100 - 5)3 = (100)3 – (5)3 – 3(100)(5)(100-5)
= 1000000 – 125 – 1500(95)
= 1000000 – 125 – 142500
= 857375.

Test: Polynomial Identities - Question 3

26 * 34 = __________ (calculate without direct calculation).

Detailed Solution for Test: Polynomial Identities - Question 3

26 * 34 can also be written as (30 - 4) * (30 + 4)
We know that (a - b) * (a + b) = a2 - b2
Similarly, 26 * 34 = (30 - 4) * (30 + 4)
= 302 – 42
= 900 – 16
= 884.

Test: Polynomial Identities - Question 4

What do we get after factorising x+ 6x - 27?

Detailed Solution for Test: Polynomial Identities - Question 4

To factorise x+ 6x - 27, we have to find two numbers ‘a’ and ‘b’ such that a + b = 6 and a * b = 27.
For that we have to find factors of -27, which are ±1, ±3, ±9.
Now we have to arrange two numbers from these numbers such that a + b = 6 and a * b = 27.
By considering this, we get two numbers +9 and -3
9 + (-3) = 6 and 9*-3 = -27
Now after manipulating terms, we get x+ 9x - 3x - 27.
x+ 9x - 3x - 27 = x(x + 9) - 3(x + 9)
= (x + 9)(x - 3).

Test: Polynomial Identities - Question 5

x - 1 is a factor of 4x- 9x - 6.

Detailed Solution for Test: Polynomial Identities - Question 5

According to factor theorem, x - a is a factor of p(x) if p(a) = 0.
Therefore x-1 is a factor of 4x- 9x - 6 is a factor if p(1) = 0.
p(1) = 4(1)- 9(1) - 6 = 4 – 9 – 6
= -11 ≠ 0
Therefore, we can say that x - 1 is not a factor of 4x- 9x - 6.

Test: Polynomial Identities - Question 6

What do we get after expanding (p + 3q - 2z)2?

Detailed Solution for Test: Polynomial Identities - Question 6

We know that (a+b+c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
(p + 3q - 2z)2can also be written as (p + 3q + (-2z))2.
Here, a = p, b = 3q and c = -2z
Therefore, using that formula we get (p + 3q + (-2z))2
= p2 + (3q)2 + (-2z)2 + 2(p)(3q) + 2(3q)(-2z) + 2(-2z)(p)
= p2 + 9q2 + 4z2 + 6pq – 12qz – 4zp.

Test: Polynomial Identities - Question 7

27 * 29 = __________ (calculate without direct calculation).

Detailed Solution for Test: Polynomial Identities - Question 7

We know that (a + b) * (a + c) = a2 + (b + c)a + bc
27 * 29 can also be written as (25 + 2) * (25 + 4)
Now using above identity, 27 * 29 = (25 + 2) * (25 + 4)
= 252 + (2 + 4)25 + (4)(2)
= 625 + 6(25) + 8
= 625 + 150 + 8
= 783.

Test: Polynomial Identities - Question 8

What do we get after factoring 49x- 28xy + .4y2?

Detailed Solution for Test: Polynomial Identities - Question 8

We know that a- 2ab + .b2 = (a - b)2
49x- 28xy + .4y2 can also be written as (7x)- 2(7)(2)xy + (2y)2
Here, a = 7x and b = 2y.
Therefore, 49x- 28xy + .4y2 = (7x)- 2(7)(2)xy + (2y)2
= (7x - 2y)2.

Test: Polynomial Identities - Question 9

Find the value of k, if x-3 is a factor of 5x- 2x+ x + k.

Detailed Solution for Test: Polynomial Identities - Question 9

According to factor theorem, x-a is a factor of p(x) if p(a) = 0.
Here, it is given that x - 3 is a factor of 5x- 2x+ x + k.
Therefore, p(3) must be equal to zero.
p(3) = 5(3)- 2(3)+ 3 + k = 0
Therefore, 5(27) – 2(9) + 3 + k = 0
135 – 18 + 3 + k = 0
120 + k = 0
Therefore, k = -120

Test: Polynomial Identities - Question 10

If we add, 7xy + 5yz – 3zx, 4yz + 9zx – 4y and –3xz + 5x – 2xy, then the answer is

Detailed Solution for Test: Polynomial Identities - Question 10

Given, 7xy + 5yz – 3zx, 4yz + 9zx – 4y and –3xz + 5x – 2xy.

If we add the three expressions, then we need to combine the like terms together.

(7xy – 2xy) + (5yz + 4yz) – 3zx + 9zx – 3xz – 4y + 5x

= 5xy + 9yz + 3zx + 5x – 4y

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