5 Marks Questions

Question 15 Marks

Divide: $44(x^4– 5x^3– 24x^2)$ by $11x(x – 8)$

Answer

Factorising $44(x^4– 5x^3– 24x^2)$, we get
$44(x^4– 5x^3– 24x^2)$
$= 2 \times 2 \times 11 \times x^2(x^2- 5x - 24)$ [taking the common factor $x^2$ out of the bracket]
$= 2 \times 2 \times 11 \times x^2(x^2– 8x + 3x – 24)$
$= 2 \times 2 \times 11 \times x^2[x(x – 8) + 3(x – 8)]$
$= 2 \times 2 \times 11 \times x^2 \times (x + 3)(x – 8)$
Therefore, $4(x^4– 5x^3– 24x^2) \div 11x(x - 8)$
$= \frac{2 \times 2 \times 11 \times x \times x \times(x+3) \times(x-8)}{11 \times x \times(x-8)}$
$= 2 \times 2 \times x(x + 3) = 4x(x + 3)$

View full question & answer→

Question 25 Marks

Divide $24\left(x^2 y z+x y^2 z+x y z^2\right)$ by 8xyz using both the methods.

Answer

$24\left(x^2 y z+x y^2 z+x y z^2\right)$
$= 2 \times 2 \times 2 \times 3 \times [(x \times x \times y \times z) + (x \times y \times y \times z) + (x \times y \times z \times z)]$
$= 2 \times 2 \times 2 \times 3 \times x \times y \times z \times (x + y + z)$ [By taking out the common factor]
$= 8 \times 3 \times xyz \times (x + y + z)$
Now, 2$4\left(x^2 y z+x y^2 z+x y z^2\right) \div 8xyz$
$= \frac {8\times 3 ×xyz \times (x + y + z)}{8\times xyz} = 3 \times (x + y + z) = 3(x + y + z)$
Alternately, $24\left(x^2 y z+x y^2 z+x y z^2\right) \div 8xyz$
$=\frac{24 x^{2} y z}{8 x y z}+\frac{24 x y^{2} z}{8 x y z}+\frac{24 x y z^{2}}{8 x y z}$
$= 3x + 3y + 3z = 3(x + y + z)$

View full question & answer→

Maths 5 Marks Questions

Take a timed test