### Polynomials - Test Papers

CBSE Test Paper 01

CH-2 Polynomials

1. If p(x) = x + 3, then p(x) + p(-x) is equal to
1. 2x
2. 3
3. 0
4. 6
2. If x+y+z = 0, then
1. 3xyz
2. xyz
3. 2xyz
4. 0
3. The degree of constant function is
1. 0
2. 3
3. 1
4. 2
4. (x + 1) is a factor of the polynomial
1. ${x}^{3}+{x}^{2}-x+1$
2. ${x}^{3}+{x}^{2}+x+1$
3. ${x}^{4}+3{x}^{3}+3{x}^{2}+x+1$
4. ${x}^{4}+{x}^{3}+{x}^{2}+1$
5. The coefficient of ‘x’ in the expansion of $\left(x+3{\right)}^{3}$ is
1. 1
2. 27
3. 9
4. 18
6. Fill in the blanks: A polynomial containing one non-zero term is called a ________.
7. Fill in the blanks: The highest power of the variable in a polynomial is called the ________ of the polynomial.
8. Write (-3x + y + z)2 in the expanded form:
9. Evaluate the following by using identities: 0.54 $×$ 0.54 - 0.46 $×$ 0.46
10. Evaluate: 185 $×$ 185 - 15 $×$ 15
11. Evaluate 105 $×$ 108 without multiplying directly.
12. Simplify (x + y)3 - (x - y)3 - 6y(x + y) (x - y).
13. Factorize : (a2 - b2)3 + (b2 - c2)3 + (c2 - a2)3
14. If both x - 2 and $x-\frac{1}{2}$ are factors of px2 + 5x + r, show that p = r.
15. If the polynomials az3 + 4z2 + 3z - 4 and z3 - 4z + a leave the same remainder when divided by z - 3, find the value of a.

CBSE Test Paper 01
CH-2 Polynomials

Solution

1. (d) 6
Explanation:
p(x) = x + 3
And p(-x) = -x + 3
Then, p(x) + p(-x)
= x + 3 - x + 3
= 6
2. (a) 3xyz
Explanation: ${x}^{3}+{y}^{3}+{z}^{3}-3xyz=\left(x+y+z\right)\left({x}^{2}+{y}^{2}+{z}^{2}-xy-yz-zx\right)$ =>${x}^{3}+{y}^{3}+{z}^{3}-3xyz=\left(0\right)\left({x}^{2}+{y}^{2}+{z}^{2}-xy-yz-zx\right)$ => ${x}^{3}+{y}^{3}+{z}^{3}-3xyz=0$ => ${x}^{3}+{y}^{3}+{z}^{3}=3xyz$ If x+y+z = 0, then 3xyz
3. (a) 0
Explanation: The degree of any constant term 5 (say)
We can write 5 as 5 x 1 = 5xo [Since ao= 1]
Therefore the degree of any constant term is 0
4. (b) ${x}^{3}+{x}^{2}+x+1$
Explanation: ${x}^{3}+{x}^{2}+x+1$ = ${x}^{3}\left(x+1\right)+1\left(x+1\right)$ = $\left({x}^{3}+1\right)\left(x+1\right)$
5. (b) 27
Explanation:$\left(x+3{\right)}^{3}$   = ${x}^{3}+{\left(3\right)}^{3}+3×x×3\left(x+3\right)$ = ${x}^{3}+27+9{x}^{2}+27x$ = ${x}^{3}+9{x}^{2}+27x+27$ Therefore, the coefficient of $x$, in the expansion of $\left(x+3{\right)}^{3}$ is 27.
6. monomial
7. degree
8. We have,
(-3x + y + z)2
Using identity (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
= (-3x)2 + y2 + z2 + 2 $×$ (-3x) $×$ y + 2 $×$ y $×$ z + 2 $×$ z $×$ (-3x)
= 9x2 + y2 + z2 - 6xy + 2yz - 6zx
9. We have,
0.54 $×$ 0.54 - 0.46 $×$ 0.46
= (0.54)2 - (0.46)2 = (0.54 + 0.46)(0.54 - 0.46) = 1 $×$ 0.08 = 0.08
10. 185 $×$185 - 15 $×$ 15
$\left(185{\right)}^{2}-\left(15{\right)}^{2}$ [Using ${a}^{2}-{b}^{2}=\left(a-b\right)\left(a+b\right)$]
$\left(185+15\right)\left(185-15\right)$
200 $×$ 170 = 34000
11. $105$$×$ $108=\left(100+5\right)\left(100+8\right)$
Using identity $\left(x+a\right)\left(x+b\right)={x}^{2}+\left(a+b\right)x+ab$
We get, 105 $×$ 108 = 1002 + (5 + 8)100 + 5 $×$ 8
$=10000+1300+40=11340$
12. (x + y)- (x - y)3 - 6y(x + y)(x - y)
= (x + y)3 - (x - y)3 - 3.2y(x + y)(x - y)
= (x + y)3 - (x - y)3 - 3 (x + y - x + y)(x + y)(x - y)
= [x + y - x + y)]3 [$\therefore$ a- b3 - 3ab(a - b) = (a - b)3]
=(2y)3 = 8y3
13. Let x = a2 - b2, y = b2 - c2 and z = c2 - a2. Then,
x + y + z = a2 - b2 + b2 - c2 + c2 - a2 = 0
$\therefore$ x3 + y3 + z3 = 3xyz
$⇒$ (a2 - b2)3 + (b2 - c2)3 + (c2 - a2)3 = 3(a2 - b2)(b2 - c2)(c2 - a2)
$⇒$ (a2 - b2)3 + (b2 - c2)3 + (c2 - a2)3 = 3(a + b)(a - b)(b + c)(b - c)(c + a)(c - a)
$⇒$ (a2 - b2)3 + (b2 - c2)3 + (c2 - a2)3 = 3(a + b)(b + c)(c + a)(a - b)(b - c)(c - a)
14. Let f(x) = px2 + 5x + r be the given polynomial. Since x - 2 and $x-\frac{1}{2}$are factors of f(x).
$\therefore$ f(2) = 0 and $f\left(\frac{1}{2}\right)$ = 0
$⇒$ p $×$ 2+ 5 $×$ 2 + r = 0 and p($\frac{1}{2}$)+ 5 $×\frac{1}{2}$ + r = 0
$⇒$ 4p + 10 + r = 0 and $\frac{p}{4}+\frac{5}{2}$ + r = 0
$⇒$ 4p + r = -10 and $\frac{p+4r+10}{4}=0$
$⇒$ 4p + r = -10 and p + 4r + 10 = 0
$⇒$ 4p + r = -10 and p + 4r = -10
$⇒$ 4p + r = p + 4r [RHS of the two equations are equal]
$⇒$ 3p = 3r $⇒$ p = r
15. Let p(z) = az3 + 4z2 + 3z - 4
And q(z) = z3 - 4z + a
As these two polynomials leave the same remainder, when divided by z - 3, then p(3) = q(3).
$\therefore$ p(3) = a(3)3 + 4(3)2 + 3(3) - 4
= 27a + 36 + 9 - 4
Or p(3) = 27a + 41
And q(3) = (3)3 - 4(3) + a
= 27 - 12 + a = 15 + a
Now, p(3) = q(3)
$⇒$ 27a + 41 = 15 + a
$⇒$ 26a = -26; a = -1
Hence, the required value of a = -1.