Test: Polynomial Identities - Grade 10 MCQ

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 5x³-2x²+x+k.
Therefore, p(3) must be equal to zero.
p(3) = 5(3)³-2(3)²+3+k = 0
Therefore, 5(27) – 2(9) + 3 + k = 0
135 – 18 + 3 + k=0
120 + k = 0
Therefore, k= -120

We know that a² - 2ab + .b² = (a-b)²
49x² - 28xy +.4y² can also be written as (7x)² - 2(7)(2)xy + (2y)²
Here, a = 7x and b = 2y.
Therefore, 49x² - 28xy+.4y² = (7x)² - 2(7)(2)xy + (2y)²
= (7x - 2y)².

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.

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.

According to factor theorem, x-a is a factor of p(x) if p(a) = 0.
Therefore x-1 is a factor of 4x² - 9x - 6 is a factor if p(1) = 0.
p(1) = 4(1)²-9(1)-6 = 4 – 9 – 6
= -11 ≠ 0
Therefore, we can say that x-1 is not a factor of 4x² - 9x - 6.

To factorise x² + 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 x² + 9x - 3x - 27.
x² + 9x - 3x - 27 = x(x + 9)-3(x + 9)
= (x+9)(x-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.

We know that (x-y)³ = x³– y³ – 3 xy (x-y).
95³ can also be written as (100-5)³
Therefore, 95³ = (100-5)³ = (100)³ – (5)³ – 3(100)(5)(100-5)
= 1000000 – 125 – 1500(95)
= 1000000 – 125 – 142500
= 857375.

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).

12x + 36
12x + 12 (3)
= 12(x + 3)