Mathematics Class IX Questions Bank

Question 1 of 67

1. Question
1 point(s)
Write whether the following statements are true of false : In each case justify your answer .
(i) $\dfrac{1}{\sqrt5}x^{\frac{1}{2}}+1 $ is a polynomial
(ii) $\dfrac{6\sqrt x+x^{\frac{3}{2}}}{\sqrt x} $ is a polynomial, $x\ne0 $.
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Hint

(i) False, because its power $\dfrac{1}{2 }$ is not on integer
(ii) $\dfrac{6x^{\frac{1}{2}}+x^{\frac{3}{2}}}{x^{\frac{1}{2}}}=\dfrac{x^{\frac{1}{2}}(6+x^1)}{x^{\frac{1}{2}}}$
$=6+x^1$ is a polynomial as at power is an integer
Question 2 of 67

2. Question
1 point(s)
Find the zeroes of the polynomial $p(x) = x^2-5x+6. $
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Hint

$x^2-5x+6\\ \\ x^2-3x-2x+6\\ \\ x(x-3)-2(x-3)\\ \\ (x-3)(x-2)$
The zeroes are 2 and 3
Question 3 of 67

3. Question
1 point(s)
For the polynomial $\dfrac{x^3+2x+1}{5}-\dfrac{7}{2}x^2-x^5 $, write
(i) The degree of the polynomial
(ii) The coefficient of $x^3 $
(iii) The coefficient of $x^6 $
(iv) The constant term
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Hint

$\dfrac{x^3+2x+1}{5}-\dfrac{7}{2}x^2-x^5\\ \\ =\dfrac{2x^3+4x+2-35x^2-10x^5}{10}\\ \\ =\dfrac{-10x^5+2x^3-35x^2+4x+2}{10=-x^5+\dfrac{x^3}{5}-\dfrac{35x^2}{10}+\dfrac{4x}{10}+\dfrac{2}{10}}$
(i) The degree of the polynomial 5
(ii) The coefficient of $x^3$ is $\dfrac{1}{5}$
(iii) The coefficient of $x^6$ is $0$
(iv) The constant term is $\dfrac{2}{10}=\dfrac{1}{5}$
Question 4 of 67

4. Question
1 point(s)
Give an example of a polynomial which is :
(i) Monomial of degree 1
(ii) Binomial of degree 20
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Hint

(i) The example of monomial of degree $1$ is $5y$
(ii) $y^{20}+x^1$
Question 5 of 67

5. Question
1 point(s)
Show that $x+3$ is a factor of $69 + 11x – x^2 + x^3$.
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Hint

$69+11x-x^2+x^3\\ \\ =69+11\times(-3)-(-3)^2+(-3)^3\\ \\ =69-33-9-27\\ \\ =69-69=0$
Question 6 of 67

6. Question
1 point(s)
Verify that 1 is not a zero of the polynomial $4y^4 – 3y^3 + 2y^2 – 5y + 1 $.
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Hint

Let $P(y)=4y^4-3y^3+2y^2-5y+1\\ \\ P(1)=4(1)^4-3(1)^3+2(1)^2-5(1)+1\\ \\ =4-3+2-5+1\\ \\ =4-1-5+1\\ \\ =3-5+1=-2+1=-1$
So 1 is not a zero of the polynomial
Question 7 of 67

7. Question
1 point(s)
Evalute using suitable identity $(999)^3 $.
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Hint

$(999)^3=(1000-1)^3\\ \\ =(1000)^3-1^3-3\times1000\times1(1000-1)\\ \\ =1000000000-1-3000\times999\\ \\ =1000000000-1-2997000\\ \\ 1000000000-2997001\\ \\ =997002999$
Question 8 of 67

8. Question
1 point(s)
Factorise : $x^2+\dfrac{x}{4}-\dfrac{1}{8} $.
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Hint

$x^2+\dfrac{x}{4}-\dfrac{1}{8}\\ \\ x^2+\dfrac{x}{2}-\dfrac{x}{4}-\dfrac{1}{8}\\ \\ =x\left(x+\dfrac{1}{2}\right)-\dfrac{1}{4}\left(x+\dfrac{1}{2}\right)\\ \\ =\left(x-\dfrac{1}{4}\right)\left(x+\dfrac{1}{2}\right)$
Question 9 of 67

9. Question
1 point(s)
Factorise : $27p^3-\dfrac{1}{216}-\dfrac{9}{2}p^2+\dfrac{1}{4}p $.
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Hint

$27p^3-\dfrac{1}{216}-\dfrac{9p^2}{2}+\dfrac{1}{4}p\\ \\ =(3p)^3-\left(\dfrac{1}{6}\right)^3-3\times9p^2\times\dfrac{1}{6}+3\times 3p\times\dfrac{1}{36}\\ \\ =\left(3p-\dfrac{1}{6}\right)^3$
Question 10 of 67

10. Question
1 point(s)
If $x^2+\dfrac{1}{x^2}=38 $. then find the value of $\left(x-\dfrac{1}{x}\right) $.
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Hint

$x^2+\dfrac{1}{x^2}=38\\ \\ \left(x-\dfrac{1}{x}\right)^2=x^2+\dfrac{1}{x^2}-2\times x\times\dfrac{1}{x}\\ \\ =38-2=36\\ \\ x-\dfrac{1}{x}=\pm6$
Question 11 of 67

11. Question
1 point(s)
Find the product of $\left(x-\dfrac{1}{x}\right),\left(x+\dfrac{1}{x}\right),\left(x^2+\dfrac{1}{x^2}\right) $ and $\left(x^4+\dfrac{1}{x^4}\right) $.
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Hint

$\left(x-\dfrac{1}{x}\right)\left(x+\dfrac{1}{x}\right)\left(x^2+\dfrac{1}{x^2}\right)\left(x^4+\dfrac{1}{x^4}\right)\\ \\ =\left(x^2-\dfrac{1}{x^2}\right)\left(x^2+\dfrac{1}{x^2}\right)\left(x^4+\dfrac{1}{x^4}\right)\\ \\ =\left(x^4-\dfrac{1}{x^4}\right)\left(x^4+\dfrac{1}{x^4}\right)\\ \\ =(x^4)^2-\left(\dfrac{1}{x^4}\right)^2\\ \\ =x^8-\dfrac{1}{x^8}$
Question 12 of 67

12. Question
1 point(s)
Check whether the polynomial $p(x) = 3s^3+s^2-20s+12 $ is a multiple of $3s-2 $.
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Hint

$3S-2=0\ \Rightarrow \ S=\dfrac{2}{3}\\ \\ 3S^3+S^2-20S+12\\ \\ =3\left(\dfrac{2}{3}\right)^3+\left(\dfrac{2}{3}\right)^2-20\times\dfrac{2}{3}+12\\ \\ =3\times\dfrac{8}{27}+\dfrac{4}{9}-\dfrac{40}{3}+12\\ \\ =\dfrac{8}{9}+\dfrac{4}{9}-\dfrac{40}{3}+12\\ \\ =\dfrac{8+4-120+108}{9}=\dfrac{12-12}{9}=0$
Question 13 of 67

13. Question
1 point(s)
Check whether $p(x)$ is a multiple of $g(x)$ or not, where $ p(x)=x^3-x+1,\ g(x) = 2 – 3x.$
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Hint

$g(x)=2-3x=0\\ \\ \qquad 2=3x\ \Rightarrow\ x=\dfrac{2}{3}\\ \\ p\left(\dfrac{2}{3}\right)=\left(\dfrac{2}{3}\right)^3-\dfrac{2}{3}+1=\dfrac{8}{27}-\dfrac{2}{3}+1\\ \\ =\dfrac{8-18+27}{27}= \dfrac{35-18}{27}=\dfrac{17}{27}$
So $p(x)$ is not a multiple of $g(x)$
Question 14 of 67

14. Question
1 point(s)
Check whether $g(x)$ is a factor of $p(x) $ or not, where: $p(x) = 8x^3 – 6x^2 – 4x + 3,g(x) =\dfrac{x}{3}-\dfrac{1}{4} $
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Hint

$g(x)=\dfrac{x}{3}-\dfrac{1}{4}=0\\ \\ \dfrac{x}{3}=\dfrac{1}{4}\Rightarrow x=\dfrac{3}{4}\\ \\ p\left(\dfrac{3}{4}\right)=8\left(\dfrac{3}{4}\right)^3-6\times\dfrac{3}{4}-4\times\dfrac{3}{4}+3\\ \\ =8\times\dfrac{27}{64}-\dfrac{9}{2}-3+3\\ \\ =\dfrac{27}{8}-\dfrac{9}{2}=\dfrac{27-36}{8}\\ \\ =\dfrac{-9}{8}\ne0$
So $g(x)$ is not a factor of $p(x)$
Question 15 of 67

15. Question
1 point(s)
What must be added to $x^3-3x^2-12x+19 $ so that the result is exactly divisible by $x^2+x-6 $?
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Hint

So $(-2x-5)$ must be added
Question 16 of 67

16. Question
1 point(s)
Using suitable identity, evaluate the following $\mathrm{(i)\ 103^3\ (ii)\ 101 \times 102\ (C)\ 999^2}$
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Hint

$\text{(i)}\ 103^3=(100+3)^3\\ \\ =(100)^3+3^3+3\times100\times3(100+3)\\ \\ =1000000000+27+900\times103\\ \\ =1000000027+92700\\ \\ =1092727\\ \\ \text{(ii)}\ 101\times102=(100+1)\ (100+2)\\ \\ =10000+(1+2)\times100+1\\ \\ =10000+300+2\\ \\ =10302\text{C}\ 999^2=(1000-1)^2\\ \\ =1000000+1-2\times1000\\ \\ =1000001-2000\\ \\ =998001$
Question 17 of 67

17. Question
1 point(s)
Find the following product : $(2x-y+3z)\ (4x^2+y^2+9z^2+2xy+3yz-6xz) $.
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Hint

$(2x-y+3z)(4x^2+y^2+9z^2+2xy+3yz-6xz)\\ \\ =(2x-y+3z)[(2x)^2+y^2+(3z)^2-(2x\times y)-(y\times3z)-(3z\times 2x)]\\ \\ (\because(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=a^3+b^3+c^3-3abc)\\ \\ =(2x)^3+(-y)^3+(3z)^3-3(2x\times(-y)\times 3z)\\ \\ =8x^3-y^3+27z^3+18xyz$
Question 18 of 67

18. Question
1 point(s)
Find the value of $x^3+y^3-12xy+64 $ when $x+y=-4 $.
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Hint

$x^3+y^3-12xy+64\\ \\ x^3+y^3+4^3-3(4xy)\\ \\ (x+y+4)(x^2+y^2+4^2-xy-4y-4x)\\ \\ =(-4+4)(x^2+y^2+4^2-xy-4y-4x)\\ \\ =0\ (\because\ x+y=-4)$
Question 19 of 67

19. Question
1 point(s)
If $2a=3+2b $, prove that $8a^3-8b^3-36ab=27 $
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Hint

$2a=3+2b$
Cubing both the sides
$8a^3=(3+2b)^3\\ \\ 8a^3=3^3+(2b)^3+3\times3\times2b(3+2b)\\ \\ 8a^3=27+8b^3+18b\times 2a\\ \\ 8a^3=27+8b^3+36ab\\ \\ \Rightarrow8a^3-8b^3-36ab=27\ \text{(Proved)}$
Question 20 of 67

20. Question
1 point(s)
The polynomials $kx^3+3x^2-8 $ and $3x^3-5x+k $ are divided by $x+2 $. If the remainder in each case is the same, find the value of $k $.
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Hint

$kx^3+3x^2-8\\ \\ p(-2)=k(-2)^3+3(-2)^2-8\\ \\ =-8k+4\\ \\ 9(x)=3x^3-5x+k\\ \\ 9(-2)=3(-2)^3-5(-2)+k\\ \\ =-14+k$
A/Q
$-8k+4=-14+k\\ \\ \Rightarrow -9k=-18\Rightarrow k=2$
Question 21 of 67

21. Question
1 point(s)
Factorise : $4(x^2+1)^2+13(x^2+1)-12 $
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Hint

$4(x^2+1)^2+13(x^2+1)-12\\ \\ =4(x^4+2x^2+1)+13x^2+13-12\\ \\ =4x^4+8x^2+4+13x^2+1\\ \\ =4x^4+21x^2+5\\ \\ =4x^4+20x^2+x^2+5\\ \\ =4x^2(x^2+5)+1(x^2+5)\\ \\ =(4x^2+1)\ (x^2+5)$
Question 22 of 67

22. Question
1 point(s)
Factorise : $x^2+\dfrac{1}{x^2}+2-2x-\dfrac{2}{x} $.
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Hint

$x^2+\dfrac{1}{x^2}+2-2x-\dfrac{2}{x}\\ \\ =x^2+\dfrac{1}{x^2}+2\times x\times\dfrac{1}{x}-2x-\dfrac{2}{x}\\ \\ =\left(x+\dfrac{1}{x}\right)^2-2\left(x+\dfrac{1}{x}\right)\\ \\ =\left(x+\dfrac{1}{x}\right)\ \left(x+\dfrac{1}{x}-2\right)$
Question 23 of 67

23. Question
1 point(s)
Simplify : $\left(x-\dfrac{2}{3}y\right)^3-\left(x+\dfrac{2}{3}y\right)^3 $.
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Hint

$\left(x-\dfrac{2}{3}y\right)^3-\left(x+\dfrac{2}{3}y\right)^3\\ \\ =x^3-\left(\dfrac{2}{3}y\right)^3-3\times x\times \dfrac{2}{3}y\left(x-\dfrac{2}{3}y\right)-\left(x^3+\left(\dfrac{2}{3}y\right)^3+3\times x\times\dfrac{2}{3}y\left(x+\dfrac{2}{3}y\right)\right)\\ \\ =x^3-\dfrac{8}{27}y^3-2x^2y+\dfrac{4}{3}xy^2-x^3-\dfrac{8}{27}y^3\\ \\ -2x^2y-\dfrac{4}{3}xy^2\\ \\ =\dfrac{-16}{27}y^3-4x^2y$
Question 24 of 67

24. Question
1 point(s)
Using suitable identity evaluate : $(42)^3 – (18)^3 – (24)^3 $
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Hint

$(42)^3-(18)^3-(24)^3$
Let $x=42,y=-18,z=-24$
$x+y+z=42-18-24=0\\ \\ x^3+y^3+z^3=3xyz\\ \\ =3\times42\times(-18)\times(-24)\\ \\ =54432$
Question 25 of 67

25. Question
1 point(s)
Factorise $9x^2+y^2+z^2-6xy+2yz-6zx $. Hence find the value if $x=1,y=2 $ and $z=-1 $.
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Hint

$9x^2+y^2+z^2-6xy+2yz-6zx\\ \\ =(3x)^2+y^2+z^2-2\times3x\times y+2\times y\times z-2\times z\times3x\\ \\ =(3x-y-z)^2\\ \\ x=1,y=2,z=-1\\ \\ (3x-y-z)^2\\ \\ =[3(1)-2-(-1)]^2\\ \\ =(3-2+1)^2=2^2=4$
Question 26 of 67

26. Question
1 point(s)
If $a + b + c = 6, $ then find the value of $(2-a)^3 +(2-b)^3 +(2-c)^3-3(2-a)(2-b)(2-c). $
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Hint

$a+b+c=6\\ \\ (2-a)^3+(2-b)^3+(2-c)^3-3(2-a)(2-b)(2-c)$
Let $x=2-a,y=2-b,z=2-c\\ \\ x+y+z=(2+2+2)-(a+b+c)\\ \\ =6-6=0\\ \\ x^3+y^3+z^3-3xyz\\ \\ =(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\\ \\ =(2-a+2-b+2-c)\left\{(2-a)^2+(2-b)^2+(2-c)^2-(2-a)\ (2-b)-(2-b)\ (2-c)-(2-c)\ (2-a)\right\}\\ \\ =\left\{6-6\right\}(2-a)^2+(2-b)^2+(2-c)^2-(2-a)\ (2-b)-(2-b)(2-c)-(2-c)\ (2-a)\\ \\ =0$
Question 27 of 67

27. Question
1 point(s)
If $a^2+b^2+c^2=250 $ and $ab+bc+ca=3 $. Find $ a+b+c$.
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Hint

$(a+b+c)^2\\ \\ =a^2+b^2+c^2+2(ab+bc+ca)\\ \\ =250+2\times3\\ \\ =256\\ \\ a+b+c=\pm16$
Question 28 of 67

28. Question
1 point(s)
If $ax^3+bx^2+x-6 $ has $(x+2) $ as a factor and leaves a remainder 4 when divided by $x-2 $, find the value of $a $ and $b $.
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Hint

$P(-2)=0\\ \\ a(-2)^3+b(-2)^2-2-6=0\\ \\ =-8a+4b-8=0\\ \\ \Rightarrow 2a-b+2=0$ ….(1)
$P(2)=4\\ \\ a(2)^3+b(2)^2+2-6=4\\ \\ 2a+b-2=0$ ….(2)
Solve equation (1) and (2)
$4a=0\Rightarrow a=0$ and $b=2$
Question 29 of 67

29. Question
1 point(s)
If $p=4-q $, prove that $p^3+q^3+12pq=64 $
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Hint

$p=4-q\\ \\ \text{LHS}=p^3+q^3+12pq\\ \\ =(4-q)^3+q^3+12(4-q)q\\ \\ =64-q^3-3\times4^2\times q+3\times4\times q^2+q^3+48q-12q^2\\ \\ =64-48q+12q^2+48q-12q^2\\ \\ =64=\text{RHS}$
Question 30 of 67

30. Question
1 point(s)
What are the possible expressions for the dimensions of the cuboids whose volume is given below ?
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Hint

$\text{Volume}=12\ kg^2+8\ kg-20\ k\\ \\ 4k(3y^2+2y-5)\\ \\ =4k(3y^2+5y-3y-5)\\ \\ =4k(3y+5)\ (y-1)\\ \\ \text{Length}=4k\\ \\ \text{Breadth} =3y+5\\ \\ \text{height}=y-1$
Question 31 of 67

31. Question
1 point(s)
Without finding the cubes, find the value of : $\left(\dfrac{1}{4}\right)^3+\left(\dfrac{1}{3}\right)^3-\left(\dfrac{7}{12}\right)^3 $.
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Hint

$\left(\dfrac{1}{4}\right)^3+\left(\dfrac{1}{3}\right)^3-\left(\dfrac{7}{12}\right)^3\\ \\ \dfrac{1}{4}+\dfrac{1}{3}-\dfrac{7}{12}=0\\ \\ 3\times\dfrac{1}{4}\times\dfrac{1}{3}\times\left(\dfrac{-7}{12}\right)\\ \\ \dfrac{-7}{48}$
Question 32 of 67

32. Question
1 point(s)
Factorise : $2\sqrt2a^3+8b^3-27c^3+18\sqrt2abc $.
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Hint

$2\sqrt2\ a^3+8b^3-27c^3+18\sqrt2abc\\ \\ (\sqrt2a)^3+(2b)^3+(-3c)^3-3\times\sqrt2a\times2b\times(-3c)\\ \\ =(\sqrt2a+2b-3c)\ (2a^2+4b^2+9c^2-2\sqrt2ab+6bc+3\sqrt2ca)$
( $\because$ Using the identity $a^3+b^3+c^3-3abc=(a+b+c)\ (a^2+b^2+c^2-ab-bc-ca)$
Question 33 of 67

33. Question
1 point(s)
Factorise : $x^6-y^6 $.
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Hint

$x^6-y^6\\ \\ =(x^3)^2-(y^3)^2\\ \\ =(x^3+y^3)\ (x^3-y^3)\\ \\ =(x+y)\ (x^2-xy+y^2)(x-y)\ (x^2+xy+y^2)$
Question 34 of 67

34. Question
1 point(s)
If both $(x-2) $ and $\left(x-\dfrac{1}{2}\right) $ are factors of $px^2+5x+r $, show that $p=r $.
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Hint

$f(x)=px^2+5x+r\\ \\ f(2)=p(2)^2+5\times2+r=0\\ \\ 4p+10+r=0$ ….(1)
$f\left(\dfrac{1}{2}\right)=p\left(\dfrac{1}{2}\right)^2+5\left(\dfrac{1}{2}\right)+r=0\\ \\ \dfrac{p}{4}+\dfrac{5}{2}+r=0\\ \\ p+10+4r=0$ … (2)
Solving equation (1) and (2)
$4p+r=-10\\ \\ p+4r=-10$
So, $4p+r=p+4r\\ \\ 3p=3r\\ \\ \Rightarrow p=r$
Question 35 of 67

35. Question
1 point(s)
Find the value of $a $ if $(x+a) $ is a factor of $x^4-a^2\ x^2+3x-a $
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Hint

$x^4-a^2x^2+3x-a\\ \\ x=-a\\ \\ \Rightarrow (-a)^4-a^2(-a)^2+3(-a)-a=0\\ \\ \Rightarrow a^4-a^4-3a-a=0\\ \\ \Rightarrow -4a=0\\ \\ \Rightarrow a=0$
Question 36 of 67

36. Question
1 point(s)
Factorise by spliting the middle term : $9(x-2y)^2-4(x-2y)-13 $.
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$9(x-2y)^2-4(x-2y)-13$
Let $x-2y=a\\ \\ 9a^2-4a-13\\ \\ =9a^2+9a-13a-13\\ \\ =9a(a+1)-13(a+1)\\ \\ (9a-13)\ (a+1)\\ \\ =[9(x-2y)-13]\ [x-2y+1]\\ \\ (9x-18y-13)\ (x-2y+1)$
Question 37 of 67

37. Question
1 point(s)
Without actual division prove that $(x-2) $ is a factor of the polynomial $3x^3-13x^2+8x+12 $ Also, factorise it completely.
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Question 38 of 67

38. Question
1 point(s)
If $a,b,c $ are all non-zero and $a+b+c=0 $, prove that $\dfrac{a^2}{bc}+\dfrac{b^2}{bc}+\dfrac{c^2}{ab}=3 $
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$\dfrac{a^2}{bc}+\dfrac{b^2}{bc}+\dfrac{c^2}{ab}=3$
Multiplying by $abc$
$a^3+b^3+c^3=3abc\\ \\ \Rightarrow (a+b)^3-3ab(a+b)+c^3=3abc\\ \\ a+b+c=0\\ \\ \Rightarrow a+b=-c\\ \\ (-c)^3-3ab(-c)+c^3=3abc\\ \\ \Rightarrow -c^3+3abc+c^3=3abc\\ \\ \Rightarrow 3abc=3abc\ \text{(Proved)}$
Question 39 of 67

39. Question
1 point(s)
Prove that $(a+b+c)^3-a^3-b^3-c^3=3\ (a+b)\ (b+c)\ (c+a) $.
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$(a+b+c)^3-a^3-b^3-c^3\\ \\ =[(a+b)+c]^3-a^3-b^3-c^3\\ \\ =(a+b)^3+c^3+3c(a+b)(a+b+c)\\ \\ -a^3-b^3-c^3\\ \\ a^3+b^3+3ab(a+b)+c^3+3c(a+b)\ (a+b+c)-a^3-b^3-c^3\\ \\ =3ab(a+b)+3c(a+b)(a+b+c)\\ \\ =3(a+b)[ab+c(a+b+c)]\\ \\ =3(a+b)\ [ab+ac+bc+c^2]\\ \\ =3(a+b)\ (b+c)\ (c+a)\ \text{(Proved)}$
Question 40 of 67

40. Question
1 point(s)
If $a + b + c = 5 $ and $ab + bc + ca = 10 $, then prove that $a^3+b^3+c^3-3abc=-25 $.
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$a+b+c=5,\ ab+bc+ca=10\\ \\ a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\\ \\ (a+b+c)^2=a^2+b^2+c^2+2cab+bc+ca\\ \\ 5^2-20=a^2+b^2+c^2\\ \\ \Rightarrow a^2+b^2+c^2=5\\ \\ a^3+b^3+c^3-3abc=(a+b+c)\ (a^2+b^2+c^2-ab-bc-ca)\\ \\ =5(5-10)=-25$
Question 41 of 67

41. Question
1 point(s)
Simplify : $\dfrac{(a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3}{(a-b)^3+(b-c)^3+(c-a)^3} $
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$\dfrac{(a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3}{(a-b)^3+(b-c)^3+(c-a)^3}\\ \\ \text{Let}\ x=a^2-b^2,y=b^2-c^2,z=c^2-a\\ \\ x+y+z=a^2-b^2+b^2-c^2+c^2-a^2=0\\ \\ x+y+z=3\ (a^2-b^2)(b^2-c^2)(c^2-a^2)\\ \\ =3(a+b)(a-b)(b+c)(b-c)(c+a)(c-a)$
Again Let $x=a-b,\ y=b-c,\ z=c-a$
$x+y+z=0\\ \\ \dfrac{3(a+b)(a-b)(b+c)(b-c)(c+a)(c-a)}{3(a-b)(b-c)(c-a)}\\ \\ =(a+b)\ (b+c)\ (c+a)$
Question 42 of 67

42. Question
1 point(s)
Prove that : $2x^3+2y^3+2z^3-6xyz=(x+y+z)\ [(x-y)^2+(y-z)^2+(z-x)^2] $. Hence evaluate $2(7)^3 + 2(9)^3 + 2(13)^3 – 6(7)\ (9) (13) . $
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$2x^3+2y^3+2z^3-6xyz\\ \\ 2(x^3+y^3+z^3-3xyz)$
RHS
$(x+y+z)\ [(x-y)^2+(y-z)^2+(z-x)^2]\\ \\ =(x+y+z)\ [x^2+y^2-2xy+y^2+z^2-2yz+z^2+x^2-2xz]\\ \\ =(x+y+z)\ (2x^2+2y^2+2z^2-2xy-2yz-2zx)\\ \\ =2(x+y+z)\ (x^2+y^2+z^2-xy-yz-zx)\\ \\ =2(x^3+y^3+z^3-3xyz)\\ \\ \text{LHS=RHS}\\ \\ 2(7)^3+2(9)^3+2(13)^3-6\times7\times9\times13-a=7,b=9,c=13\\ \\ 2a^3+2b^3+2c^3-6abc$
Using Identity
$(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]\\ \\ (7+9+13)[(7-9)^2+(9-13)^2+(13-7)^2]\\ \\ =29[4+16+36]\\ \\ 29\times56=1624$
Question 43 of 67

43. Question
1 point(s)
Using factor theorem show that $x^2+5x+6 $ is factor of $x^4+5x^3+9x^2+15x+18 $.
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$p(x)=x^4+5x^3+9x^2+15x+18\\ \\ x^2+5x+6=(x+3)\ (x+2)$ zeroes are $-2,-3$
$p(-2)=(-2)^4+5(-2)^3+9(-2)^2+15(-2)+18\\ \\ =16-40+36-30+18=0\\ \\ p(-3)=(-3)^4+5(-3)^3+9(-3)^2+15(-3)+18\\ \\ =81-135+81-45+18\\ \\ =180-180=0$
So $x^2+5x+6$ is a factor of $p(x)$
Question 44 of 67

44. Question
1 point(s)
Prove that $ (x+y+z)\times[(x-y)^2+(y-z)^2]=2(x^3+y^3+z^3-3xyz)$
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$p(x)=ax^3+4x^2+3x-4\\ \\ q(x)=x^3-4x+a\\ \\ p(3)=a(3)^3+4(3)^2+3\times3-4\\ \\ =27a+36+9-4\\ \\ =27a+41\\ \\ q(3)=(3)^3-4\times3+a\\ \\ =15+a$
A/Q $27a+41=15+a\\ \\ a=-1\\ \\ p(2)=(-1)(2)^3+4(2)^2+3\times2-4\\ \\ =-8+16+6-4\\ \\ =-12+22=10$
Question 45 of 67

45. Question
1 point(s)
The polynomial $p(x) =ax^3 + 4x^2 + 3x – 4 $ and $q(x) = x^3 – 4x + a $ leave the same remainder when divided by $x-3 $. Find the remainder when $p(x) $ is divided by $(x-2) $.
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Hint

$\dfrac{6-2\sqrt2x-x^2}{2-x^2}\\ \\ =\dfrac{-x^2-2\sqrt2x+6}{-x^2+2}\\ \\ =\dfrac{x^2+2\sqrt2x-6}{x^2-2}\\ \\ =\dfrac{x^2+3\sqrt2x-\sqrt2x-6}{x^2-2}\\ \\ =\dfrac{(x+3\sqrt2)(x-\sqrt2)}{(x+\sqrt2)(x-\sqrt2)}\\ \\ =\dfrac{x+3\sqrt2}{x+\sqrt2}$
Question 46 of 67

46. Question
1 point(s)
Simplify by factrorisation method : $\dfrac{6-2\sqrt2x-x^2}{2-x^2} $
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$p(x)=x^3+2x^2-5ax-8\\ \\ q(x)=x^3+ax^2-12x-6\\ \\ p(2)=2^3+2\times2^2-5a\times(2)-8\\ \\ =8-10a\\ \\ p(3)=3^3+a\times3^2-12\times3-6\\ \\ 27+9a-36-6\\ \\ =-15+9a\\ \\ q-p=10\\ \\ -15+9a-8+10a=10\\ \\ 19a=33\\ \\ a=\dfrac{33}{19}$
Question 47 of 67

47. Question
1 point(s)
The polynomials $x^3+2x^2-5ax-8 $ and $x^3+ax^2-12x-6 $ when divided by $(x-2) $ and $(x-3) $ leave reminders $p $ and $q $ respectively. If $q-p=10 $, find the value of $a $
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Hint

$(x+y)^3-(x-y)^3-6y(x^2-y^2)\\ \\ x^3+y^3+3x^2y+3xy^2-x^3+y^3+3x^2y-3xy^2-6yx^2+6y^3\\ \\ =y^3+y^3+6y^3=8y^3$
Question 48 of 67

48. Question
1 point(s)
Prove that $(x+y)^3-(x-y)^3-6y(x^2-y^2)=8y^3 $.
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$(x-a)^3+(x-b)^3+(x-6)^3-3(x-a)\\ \\ (x-b)\ (x-c)\\ \\ x-a+x-b+x-c=(x+x+x)-(a+b+c)=3x-3x=0\\ \\ (x+a+x-b+x-c)(x-a)^2+(x-b)^2+(x-c)^2-(x-a)\ (x-b)-(x-b)\ (x-c)-(x-a)\ (x-c)\\ \\ =[3x-(a+b+c)]\ [(x-a)^2+(x-b)^2+(x-c)^2-(x-a)(x-b)-(x-b)\ (x-c)-(x-a)\ (x-c)]\\ \\ =0(\because a+b+c=3x)$
Question 49 of 67

49. Question
1 point(s)
Find the value of $(x-a)^3+(x-b)^3+(x-c)^3-3(x-a)(x-b)(x-c) $, if $a + b + c = 3x. $
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$\dfrac{9-2\sqrt3a-x^2}{3-x^2}\\ \\ =\dfrac{-x^2-2\sqrt3x+9}{-x^2+3}\\ \\ =\dfrac{x^2+2\sqrt3x-9}{x^2-3}\\ \\ =\dfrac{x^2-3\sqrt3x+\sqrt3x-9}{x^2-3}\\ \\ =\dfrac{x(x-3\sqrt3)+\sqrt3(x-3\sqrt3)}{(x+\sqrt3)(x-\sqrt3)}\\ \\ =\dfrac{(x-3\sqrt3)(x+9\sqrt3)}{(x+\sqrt3)(x-\sqrt3)}\\ \\ =\dfrac{x-3\sqrt3}{x-\sqrt3}$
Question 50 of 67

50. Question
1 point(s)
Simplify by factorisation method : $\dfrac{9-2\sqrt3x-x^2}{3-x^2} $
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$p(x)=x^3-ax^2+bx+3\\ \\ p(-2)=(-2)^3-a(-2)^2+b(-2)+3=-10\\ \\ =-8-4a-2b+3=-19\\ \\ =-4a-2b=-19+5\\ \\ =-4a-2b=-14\ …(1)\\ \\ p(2)=2^3-a(2)^2+2b+3=17\\ \\ 8-4a+2b+3=17\\ \\ -4a+2b=6\ …(2)$
Adding equation (1) and (2)
$-8a=-8\\ \\ a=1\\ \\ -4a-2b=-14\\ \\ -4-2b=-14\Rightarrow -2b=-14+4\Rightarrow b=5\\ \\ a+b=1+5=6\ \text{(Proved)}$
Question 51 of 67

51. Question
1 point(s)
If $p(x)=x^3-ax^2+bx+3 $ leaves a remainder $-19 $ when divided by $(x+2) $ and a remainder $ 17$ when divided by $(x-2) $, prove that $a+b=6 $.
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$p(x)=x^3-6x^2+12x-8\\ \\ =x^3-2^3-3\times2\times x(x-2)\\ \\ =(x-2)^3$
Side $=x-2$
when side $=3\ cm$
Volume $=3^3=27\ cm^3$
$x-2=3\\ \\ x=5$
Volume $=5^3-6\times5^2+12\times5-8$
$=125-150+60-8=27$
Question 52 of 67

52. Question
1 point(s)
The volume of a cube is given by the polynomial $p(x) = x^3 – 6x^2 + 12x – 8 $. Find the possible expression for the sides of the cube. Verify the truth of your answer when the length of cube is 3 cm.
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$p(x)=x^3-6x^2+12x-8\\ \\ =x^3-2^3-3\times2\times x(x-2)\\ \\ =(x-2)^3$
Side $=x-2$
when side $=3\ cm$
Volume $=3^3=27\ cm^3$
$x-2=3\\ \\ x=5$
Volume $=5^3-6\times5^2+12\times5-8$
$=125-150+60-8=27$
Question 53 of 67

53. Question
1 point(s)
Using factor theorem, factorise the polynomial: $x^4 + 3x^3 + 2x^2 – 3x – 3. $
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Question 54 of 67

54. Question
1 point(s)
Using factor theorem, factorise the polynomial. $x^4+2x^3-7x^2-8x+12 $.
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Question 55 of 67

55. Question
1 point(s)
Without actual division, show that the polynomail $2x^4-5x^3+2x^2-x+2 $ is exactly divisible by $x^2-3x+2 $
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$x^2-3x+2\\ \\ =x^2-2x-x+2\\ \\ (x-2)\ (x-1)\\ \\ x=2,1\\ \\ p(x)=2\\ \\ 2(2)^4-5(2)^3+2(2)^2-2+2\\ \\ =32-40+8=0\\ \\ p(x)=1\\ \\ 2(1)^4-5(1)^3+2(1)^2-1+2\\ \\ =2-5+2-1+2\\ \\ =6-6=0$
Question 56 of 67

56. Question
1 point(s)
If $x $ and $y $ be two positive real numbers such that $8x^3+27y^3=730 $ and $2x^2y+3xy^2=15 $, then evaluate $2x+3y $.
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$8x^3+27y^3=730\\ \\ (2x+3y)^3=8x^3+27y^3+18\times2x^2y+18\times3xy^2\\ \\ =730+18\times15=1000\\ \\ 2x+3y=\sqrt[3]{1000}=10$
Question 57 of 67

57. Question
1 point(s)
Without actual division prove that $x^4+ 2x^3- 2x^2+ 2x – 3 $ is exactly divisible by $x^2+2x-3. $
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$x^2+2x-3\\ \\ =x^2+3x-x-3\\ \\ x(x+3)-1(x+3)\\ \\ (x-1)\ (x+3)\\ \\ p(1)=x^4+2x^3-2x^2+2x-3\\ \\ 1^4+2\times1^3-2\times1^2+2\times1-3\\ \\ =1+2-2+2-3\\ \\ =5-5=0\\ \\ p(-3)=(-3)^4+2\times(-3)^3-2\times(-3)^2+2\times(-3)-3\\ \\ =81-54-18-6-3\\ \\ =81-81=0$
Question 58 of 67

58. Question
1 point(s)
Multiply $9x^2+25y^2+15xy+12x-20y+16 $ by $3x-5y-4 $ using suitable identify.
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$(9x^2+25y^2+15xy+12x-20y+16)\times(3x-5y-4)\\ \\ =[(3x)^2+(-5y)^2+(-4)^2-(3x)\times(-5y)-(-5y)(-4)-(-4)(3x)]\ [3x+(-5y)+(-4)]\\ \\ =(3x)^3+(-5y)^3+(-4)^3-3(3x)\ (-5y)(-4)\\ \\ =27x^3-125y^3-64-180xy\\ \\ (\because a^3+b^3+c^3-3abc=(a+b+c)\ (a^2+b^2+c^2)-ab-bc-ca)$
Question 59 of 67

59. Question
1 point(s)
Without actual division show that $x^4+2x^3-2x^2+2x-3 $ is exactly divisible by $x^2 + 2x – 3. $
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$x^2+2x-3\\ \\ =x^2+3x-x-3\\ \\ x(x+3)-1(x+3)\\ \\ (x-1)\ (x+3)\\ \\ p(1)=x^4+2x^3-2x^2+2x-3\\ \\ 1^4+2\times1^3-2\times1^2+2\times1-3\\ \\ =1+2-2+2-3\\ \\ =5-5=0\\ \\ p(-3)=(-3)^4+2\times(-3)^3-2\times(-3)^2+2\times(-3)-3\\ \\ =81-54-18-6-3\\ \\ =81-81=0$
Question 60 of 67

60. Question
1 point(s)
Factorise : $27a^3+\dfrac{1}{64b^3}+\dfrac{27a^2}{4b}+\dfrac{9a}{16b^2} $.
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$27a^3+\dfrac{1}{64b^3}+\dfrac{27a^2}{4b}+\dfrac{9a}{16b^2}\\ \\ (3a)^3+\left(\dfrac{1}{4b}\right)^3+3\times3a\times\dfrac{1}{4b}\left(3a+\dfrac{1}{4b}\right)\\ \\ =\left(3a+\dfrac{1}{4b}\right)^3$
Question 61 of 67

61. Question
1 point(s)
Find the value of $a $ and $b $ so that $(x+1) $ and $(x-2) $ are factors of $(x^3+ax^2+2x+b) $.
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Hint

$x^3+ax^2+2x+b\\ \\ p(-1)=-1+a-2+b=0\\ \\ \Rightarrow -3+a+b=0\\ \\ \Rightarrow a+b-3=0$ …(1)
$p(2)=2^3+a(2)^2+2\times2+b\\ \\ \Rightarrow 8+4a+4+b=0\\ \\ \Rightarrow 4a+b+12=0$ … (2)
Subtracting equation (1) from (2) we get
$(4a+b+12)-(a+b-3)=0$
$\Rightarrow 4a-a+15=0\\ \\ \Rightarrow a=\dfrac{-15}{3}=-5\\ \\ a+b-3=0\\ \\ -5+b=3\\ \\ b=8$
Question 62 of 67

62. Question
1 point(s)
Without actual division, prove that $(2x^4 – 6x^3 + 3x^2 + 3x – 2) $ is exactly divisible by $(x^2 – 3x + 2) $
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$x^2-3x+2\\ \\ x^2-2x-x+2\\ \\ x(x-2)-1(x-2)\\ \\ (x-1)\ (x-2)\\ \\ p(x)=2x^4-6x^3+3x^2+3x-2\\ \\ p(1)=2-6+3+3-2\\ \\ =8-8=0\\ \\ p(2)=2\times(2)^4-6\times2^3+3\times2^2+3\times2-2\\ \\ =32-48+12+6-2\\ \\ =50-50=0$
Question 63 of 67

63. Question
1 point(s)
Find the value of $a $ if $(x-a) $ is a factor of $x^5-a^2x^3+2x+a+3 $, hence factorise $x^2-2ax-3 $
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$p(x)=x^5-a^2x^3+2x+a+3\\ \\ p(a)=a^5-a^2a^3+2a+a+3=0\\ \\ \Rightarrow a^5-a^5+3a=-3\\ \\ \Rightarrow a=-1\\ \\ x^2-2ax-3\\ \\ =x^2-2(-1)x-3\\ \\ =x^2+2x-3\\ \\ =x^2+3x-x-3\\ \\ x(x+3)-1(x+3)\\ \\ (x+3)\ (x-1)$
Question 64 of 67

64. Question
1 point(s)
The polynomial $ax^3+3x^2-3 $ and $2x^3-5x+a $ when divided by $x-4 $ leave the same remainder in each case. Find the value of $a $.
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$p(x)=ax^3+3x^2-3\\ \\ p(4)=a(4)^3+3\times4^2-3\\ \\ =64a+48-3\\ \\ =64a+45\\ \\ q(x)=2x^3-5x+a\\ \\ q(4)=2\times4^3-5\times4+a\\ \\ =108+a$
A/Q
$64a+45=108+a\\ \\ \Rightarrow 63a=63\\ \\ \Rightarrow a=1$
Question 65 of 67

65. Question
1 point(s)
If $x+\dfrac{1}{x}=5 $, then evaluate $x^6+\dfrac{1}{x^6} $.
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Hint

$x+\dfrac{1}{x}=5\\ \\ \left(x+\dfrac{1}{x}\right)^2=25\\ \\ \Rightarrow x^2+\dfrac{1}{x^2}+2=25\\ \\ \Rightarrow x^2+\dfrac{1}{x^2}-23=0$
If $a+b+c=0$ then $a^3+b^3+c^3=3xyz$
$(x^2)^3+\left(\dfrac{1}{x^2}\right)^3+(-23)^3=3\times x^2\times\dfrac{1}{x^2}\times(-23)\\ \\ x^6+\dfrac{1}{x^6}-12167=-69\\ \\ \Rightarrow x^6+\dfrac{1}{x^6}=12098$
Question 66 of 67

66. Question
1 point(s)
Without actual division, prove that $2x^4-8x^3+3x^2+12x-9 $ is exactly divisible by $x^2-4x+3 $.
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Hint

$x^2-4x+3\\ \\ x^2-3x-x+3\\ \\ x(x-3)-1(x-3)\\ \\ (x-1)\ (x-3)\\ \\ p(x)=2x^4-8x^3+3x^2+12x-9\\ \\ p(1)=2(1)^4-8(1)^3+3(1)^2+12(1)-9\\ \\ =2-8+3+12-9\\ \\ =17-17=0\\ \\ p(3)=2(3)^4-8(3)^3+3(3)^2+12\times3-9\\ \\ =162-216+27+36-9\\ \\ =225-225=0$
Question 67 of 67

67. Question
1 point(s)
If $f(x) = x^4 – 2x^3 + 3x^2 – ax + b $ is divided by $(x-1) $ and $(x+1) $, it leaves the remainder 5 and 19 respectively. Find $a $ and $b $.
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Hint

$f(x)=x^4-2x^3+3x^2-ax+b\\ \\ f(1)=1-2+3-a+b=5\\ \\ \Rightarrow 2-a+b=5\\ \\ \Rightarrow -a+b=3$ …(1)
$f(-1)=(-1)^4-2(-1)^3+3(-1)^2-a(-1)+b\\ \\ \Rightarrow 1+2+3+a+b=19\\ \\ \Rightarrow a+b=13$ ….(2)
Adding equation (1) and (2)
$2b=16\\ \\ \Rightarrow b=8\\ \\ -a+b=3\\ \\ \Rightarrow -a=3-b=3-8=-5\\ \\ \Rightarrow a=5$