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​Algebra 1   State Test Practice

Produce an equivalent form of an expression

1/26/2015

 
Which factorization can be used to reveal the zeros of the function  
f(n) = -12n² - 11n + 15?

A.)  f(n) = -n(12n + 11) + 15
B.)  f(n) = (-4n + 3)(3n + 5)
C.)  f(n) = -(4n + 3)(3n + 5)
D.)  f(n) = (4n + 3)(-3n + 5)

f(n) = -12n² - 11n + 15
= -1(12n² + 11n - 15)   {factored -1 out}

12n² + 11n - 15  needs to be factored by grouping:
- split the middle term, 11n, into two terms who's coefficients
multiply to get what 12(-15) is......-180
and add to get the middle coefficient 11


So, two numbers that multiply to get -180 and add to get 11 are.......20 and -9
12n² + 11n - 15
= 12n² + 20n - 9n - 15   {split the middle term, 11n, into 20n and -9n}
= 4n(3n + 5) - 3(3n + 5)   {factoring by grouping, factored 4n out of 1st two terms and -3 out of last two}
= (4n - 3)(3n + 5)   {factored out the common factor, (3n + 5)}

f(n) = -1(12n² - 11n + 15)
f(n) = -1(4n - 3)(3n + 5)   {factored by grouping}
B.)  f(n) = (-4n + 3)(3n + 5)   {distributed the negative sign through 1st set,
                                               so that it would match an answer choice}


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