3 Marks Question

Question 513 Marks

Solve:$2a^5-32a$

Answer

$2 a^5-32 a$
$=2 a\left(a^4-16\right)$
$=2 a\left[\left(a^2\right)^2-4^2\right] $
$=2 a\left(a^2+4\right)\left(a^2-4\right)$
$=2 a\left(a^2+4\right)\left(a^2-2^2\right)$
$=2 a\left(a^2+4\right)(a+2)(a-2)$
$ =2 a(a-a)(a+2)\left(a^2+4\right)$

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Question 523 Marks

Solve:$16-a^6+4 a^3 b^3-4 b^6$

Answer

$16-a^6+4 a^3 b^3-4 b^6$
$=16-\left(a^6-4 a^3 b^3+4 b^6\right)$
$ =4^2-\left[\left(a^3\right)^2-2 \times a^3 \times 2 b^3+\left(2 b^3\right)^2\right]$
$ =4^2-\left(a^3-2 b^3\right)^2$
$=\left[4-\left(a^3-2 b^3\right)\right]\left[4+\left(a^3-2 b^3\right)\right]$
$\left.=\left(4-a^3-2 b^3\right)\left(4+a^3-2 b^3\right)\right]$
$=\left(a^3-2 b^3+4\right)\left(-a^3-2 b^3+4\right)$

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Question 533 Marks

Factories:$x^2-22 x+120$

Answer

To factories $x^2-22 x+120$, we will find two number $p$ and $q$ such that $p+q=-22$ and $p q=120$
Now,
$(-12)+(-10)=-22 \text { And }(-22) \times(-10)=120$
Splittiong the middle term $14 x$ in the given quadratic as $-12 x-10 x$, we get:
$x^2-22 x+12=x^2-12 x-10 x+120$
$=\left(x^2-12\right)+(-10 x+120)$
$= x(x - 12) - 10(x - 12)$
$= (x - 10)(x - 12)$

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Question 543 Marks

Find the greatest common factor of the polynomial:$3 a^2 b^2+4 b^2 c^2+12 a^2 b^2 c^2$

Answer

The numerical coefficients of the given monomials are $3 a^2 b^2, 4 b^2 c^2$ and $12 a^2 b^2 c^2$.
The greatest common factor of $3 a^2 b^2, 4 b^2 c^2$ and $12 a^2 b^2 c^2$ is $1 .$
The common literal appearing in the three monomials is $b$ .
The smallest power of $b$ in the three monomials is $2 .$
The monomial of common literals with the smallest powers is $b ^2$.
Hence, the greatest common factor is $b^2$.

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Question 553 Marks

Find the greatest common factor of the polynomial: $2 x y z+3 x^2 y+4 y^2$

Answer

The numerical coefficients of the given monomials are $2 x y z, 3 x^2 y$ and $4 y^2$.
The greatest common factor of $2 x y z, 3 x^2 y$ and $4 y^2$ is $1 .$
The common literal appearing in the three monomials is $y$.
The smallest power of $y$ in the three monomials is $1.$
The monomial of common literals with the smallest powers is $y$.
Hence, the greatest common factor is $y$.

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Question 563 Marks

Solve: $q^2-10 p+21$

Answer

$q^2-10 p+21$
$\left.=q^2-10 q+\left(\frac{10}{2}\right)^2-\left(\frac{10}{2}\right)^2+21 \text { [Adding and subtracting }\left(\frac{10}{2}\right)^2, \text { that is } 5^2\right] $
$=q 2-2 \times q \times 5+5^2-5^2+21$
$=(q-5) 2-4[\text { completing the square }]$
$=[(q-5)-2][(q-5)+2]$
$=(q-5-2)(q-5+2)$
$=(q-7)(q-3)$

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MATHS 3 Marks Question