What is the standard form of y= (3x-7)(x-14)(x-11)y=(3x−7)(x−14)(x−11)?
1 Answer
3x^3 - 82x^2 + 637x - 10783x3−82x2+637x−1078
Explanation:
Require to distribute the brackets. Starting with the 1st pair and using FOIL.
(3x - 7 )(x - 14 ) = 3x^2 - 42x - 7x + 98 (3x−7)(x−14)=3x2−42x−7x+98 'collecting like terms' gives:
3x^2 - 49x +983x2−49x+98 This now requires to be multiplied by ( x - 11 )
(3x^2 - 49x +98 )(x - 11 ) (3x2−49x+98)(x−11)
each term in the 2nd bracket requires to be multiplied by each term in the 1st bracket. This is achieved by the following :
3x^2(x-11) - 49x(x-11) +98(x-11) 3x2(x−11)−49x(x−11)+98(x−11)
= 3x^3 - 33x^2 - 49x^2 +539x + 98x - 1078 =3x3−33x2−49x2+539x+98x−1078 writing in standard form means starting with the term with the largest exponent of x and then terms with decreasing terms of exponents.
rArr 3x^3 -82x^2 + 637x -1078⇒3x3−82x2+637x−1078
