# What is the standard form of  y= (3x-7)(x-14)(x-11)?

Feb 6, 2016

$3 {x}^{3} - 82 {x}^{2} + 637 x - 1078$

#### Explanation:

Require to distribute the brackets. Starting with the 1st pair and using FOIL.

$\left(3 x - 7\right) \left(x - 14\right) = 3 {x}^{2} - 42 x - 7 x + 98$

'collecting like terms' gives: $3 {x}^{2} - 49 x + 98$

This now requires to be multiplied by ( x - 11 )

$\left(3 {x}^{2} - 49 x + 98\right) \left(x - 11\right)$
each term in the 2nd bracket requires to be multiplied by each term in the 1st bracket. This is achieved by the following :

$3 {x}^{2} \left(x - 11\right) - 49 x \left(x - 11\right) + 98 \left(x - 11\right)$

$= 3 {x}^{3} - 33 {x}^{2} - 49 {x}^{2} + 539 x + 98 x - 1078$

writing in standard form means starting with the term with the largest exponent of x and then terms with decreasing terms of exponents.

$\Rightarrow 3 {x}^{3} - 82 {x}^{2} + 637 x - 1078$