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Question 1: Find the remainder when $x^3 + 3x^2 + 3x + 1$ is divided by $x + 1$.
Solution: Let $p(x) = x^3 + 3x^2 + 3x + 1$. By Remainder Theorem, remainder is $p(-1)$.
$p(-1) = (-1)^3 + 3(-1)^2 + 3(-1) + 1 = -1 + 3 - 3 + 1 = 0$. -
Question 2: Check whether $7 + 3x$ is a factor of $3x^3 + 7x$.
Solution: Put $3x + 7 = 0 \Rightarrow x = -7/3$.
$p(-7/3) = 3(-7/3)^3 + 7(-7/3) = 3(-343/27) - 49/3 = -343/9 - 147/9 \neq 0$.
No, it is not a factor. -
Question 3: Factorise $12x^2 - 7x + 1$.
Solution: Splitting the middle term: $12x^2 - 4x - 3x + 1$
$= 4x(3x - 1) - 1(3x - 1) = (4x - 1)(3x - 1)$. -
Question 4: Factorise $x^3 - 2x^2 - x + 2$.
Solution: Grouping terms: $x^2(x - 2) - 1(x - 2)$
$= (x^2 - 1)(x - 2) = (x - 1)(x + 1)(x - 2)$. -
Question 5: Expand $(2x - y + z)^2$.
Solution: Using $(a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca$:
$= (2x)^2 + (-y)^2 + z^2 + 2(2x)(-y) + 2(-y)(z) + 2(z)(2x)$
$= 4x^2 + y^2 + z^2 - 4xy - 2yz + 4zx$. -
Question 6: Evaluate $(99)^3$ using a suitable identity.
Solution: $(100 - 1)^3 = 100^3 - 1^3 - 3(100)(1)(100 - 1)$
$= 1000000 - 1 - 300(99) = 1000000 - 1 - 29700 = 970299$. -
Question 7: Factorise $8x^3 + 27y^3 + 36x^2y + 54xy^2$.
Solution: This is of the form $(a+b)^3 = a^3 + b^3 + 3a^2b + 3ab^2$.
$= (2x)^3 + (3y)^3 + 3(2x)^2(3y) + 3(2x)(3y)^2 = (2x + 3y)^3$. -
Question 8: Factorise $27y^3 + 125z^3$.
Solution: Using $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$:
$= (3y)^3 + (5z)^3 = (3y + 5z)((3y)^2 - (3y)(5z) + (5z)^2)$
$= (3y + 5z)(9y^2 - 15yz + 25z^2)$. -
Question 9: Without calculating cubes, find the value of $(-12)^3 + (7)^3 + (5)^3$.
Solution: Let $a=-12, b=7, c=5$. Since $a+b+c = -12+7+5 = 0$,
$a^3+b^3+c^3 = 3abc = 3(-12)(7)(5) = -1260$. -
Question 10: If $x + y + z = 0$, show that $x^3 + y^3 + z^3 = 3xyz$.
Solution: We know $x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)$.
Since $x+y+z=0$, RHS becomes 0.
$\therefore x^3+y^3+z^3-3xyz = 0 \Rightarrow x^3+y^3+z^3 = 3xyz$.