4 Marks Questions

Question 14 Marks

Prove that $(a + b + c)^3 - a^3 - b^3 - c^3 = 3(a + b )(b + c)(c + a).$

Answer

$(a + b + c)^3 = [a + (b + c)]^3$
$= a^3 + 3a^2(b + c) + 3a(b + c)^2 + (b + c)^3$
$= a^3 + 3a^2b + 3a^2c + 3a(b^2 + 2bc + c^2) + (b^3 + 3b^2c + 3bc^2 + c^3)$
$= a^3 + 3a^2b + 3a^2c + 3ab^2 + 6abc + 3ac^2 + b^3 + 3b^2c + 3bc^2 + c^3$
$= a^3 + b^3 + c^3 + 3a^2b + 3a^2c + 3b^2c + 3c^2a + 3c^2b + 6abc$
$= a^3 + b^3 + c^3 + 3a^2(b + c) + a^3 + b^3 + c^3 + 3a^2(b + c)$
Hence, above result can be put in the form $(a + b + c)^{3 }= (a + b + c)^3 + 3(a + b)(b + c)(c + a)$
$\therefore (a + b + c)^3 - a^3 - b^3 - c^3 = 3(a + b)(b + c)(c + a)$

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Question 24 Marks

Multiply$ x^2 + 4y^2 + z^2 + 2xy + xz - 2yz$ by $(-z + x - 2y).$

Answer

$(x^2 + 4y^2 + z^2 + 2xy + xz - 2yz)(-z + x - 2y)$
$= x^2(-z + x - 2y) + 4y^2(-z + x - 2y) + z^2(-z + x - 2y) + 2xy(-z + x - 2y) $$+ xz(-z + x - 2y) - 2yz(-z + x - 2y)$
$ = x^2z + x^3 - 2x^2y - 4y^2z + 4xy^2 - 8y^3 - z^3 + $$xz^2 - 2yz^2 - 2xyz + 2x^2y - 4xy^2 - xz^2 + x^2z - 2xyz + 2yz^2 - 2xyz + 4y^2z$
$ = (-x^2z + x^2z ) + x^3 + (-2x^2y + 2x^2y) + (-4y^2z + 4y^2z) + $$(4xy^{2 }- 4xy^2)- 8y^3 - z^3 + (xz^2 - xz^2) + (-2yz^{2 }+ 2yz^2) + (-2xyz - 2xyz - 2xyz)$
$ = x^3 - 8y^3 - z^3 - 6xyz$
Alternate Answer:
Now$, (x - 2y - z)(x^2 + 4y^2 + z^2 + 2xy + xz - 2yz)$
$= (x - 2y - z)[(x)^2 + (-2y)^2 + (-z)^2 - (x)(-2y) - (-2y)(-z) - (x)(-z)]$
$= (x^3) + (-2y)^3 + (-z)^3 - 3(x)(-2y)(-z) $
$[$Using identity, $a^3 + b^{3 }+ c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)]$
$= x^3 - 8y^3 - z^3 - 6xyz$

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Question 34 Marks

Without actual division, prove that $2x^4 - 5x^3 + 2x^2 - x + 2$ is divisible by $x^2 - 3x + 2.$
$[$Hint: Factorise $x^2 - 3x + 2]$

Answer

We have,
$x^2 - 3x + 2 = x^2 - x - 2x + 2$
$= x(x - 1) - 2(x - 1)$
$= (x - 1)(x - 2)$
Let $f(x) = 2x^4 - 5x^3 + 2x^2 - x + 2$
Now, $p(1) = 2(1)^4 - 5(1)^3 + 2(1)^2 - 1 + 2 = 2 - 5 + 2 - 1 + 2 = 0$
$p(1) = 0$
Therefore$, (x - 1)$ divides $p(x)$
And $p(2) = 2(2)^4 - 5(2)^3 + 2(2)^2 - 2 + 2$
$= 32 - 40 + 8 - 2 + 2 = 0$
$p(2) = 0$
Therefore$, (x - 2)$ divides $p(x).$
So$, (x - 1)(x - 2) = x^2 - 3x + 2$ divides $2x^2 - 5x^3 + 2x^2 - x + 2$

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Question 44 Marks

The polynomial $p(x) = x^4 - 2x^3 + 3x^2 - ax + 3a - 7$ when divided by $x + 1$ leaves the remainder $19.$ Find the values of a. Also find the remainder when $p(x)$ is divided by $x + 2.$

Answer

We know that if $p(x)$ is divided by $x + a,$ then the remainder $= p(-a).$
Now, $p(x) = x^4 - 2x^3 + 3x^2 - ax + 3a - 7$ is divided by $x + 1,$ then the remainder $= p(-1)$
Now,$ p(-1) = (-1)^4 - 2(-1)^3 + 3(-1)^2 - a(-1) + 3a - 7$
$= 1 - 2(-1) + 3(1) + a + 3a - 7$
$= 1 + 2 + 3 + 4a -7$
$= -1 + 4a$
Also, remainder $= 19$
$\therefore -1 + 4a = 19$
$\Rightarrow 4a = 20; a = 20 \div 4 = 5$
Again, when $p(x)$ is divided by $x + 2,$ then
Remainder $= p(-2) = (-2)^4 - 2(-2)^3 + 3(-2)^2 - a(-2) + 3a -7$
$= 16 + 16 + 12 + 2a + 3a -7$
$= 37 + 5a$
$= 37 + 5(5) = 37 + 25 = 62$

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