Precalculus help answer asap

Question

(a²-a-12)/(a²-9) * (a²-4a+3)/(a^2-4a)

 

Thanks

Answer 1

This requires some Algebra...

 

First note that when we have a product of 2 fractions x/y * w/z, we can multiply numerators and denominators to get one single fraction (only dividing once).

 

x/y * w/z = xw/yz

 

Thus:

 

(a2-a-12)/(a2-9) * (a2-4a+3)/(a2-4a) = (a2-a-12)(a2-4a+3)/(a2-9)(a2-4a)

 

Next we need to use the distributive law to recognize how to multiply polynomials...when we are multiplying binomials of the form (x+y)...(x+y)(w+z) we use the FOIL mnemonic device to remember we get four terms from the distributive law...that is:

 

(x+y)(w+z) = xw+xz+yw+yz

xw = product of the FIRST terms

xz = product of the OUTER terms

yw = product of the INNER terms

yz = product of the LAST terms

F.O.I.L. = FIRST OUTER INNER LAST

 

Putting this together to find the new denominator of our new expression:

(a2-9)(a2-4a) = a^2*a^2 + a^2*(-4a) + (-9)*a^2 + (-9)(-4a)

 

The numerator is similar, but there are many more terms (9 to be exact)...

we use the distributive law again to see that (A+B+C)(X+Y+Z) = AX+AY+AZ+BX+BY+BZ+CX+CY+CZ

 

Substituting should allow you to expand out the numerator of the new expression...

 

Next we need to use the rules for combining exponents...x^A*x^B = x^(A+B)...

 

For the denominator this lets us simplify as follows:

a^2*a^2 + a^2*(-4a) + (-9)*a^2 + (-9)(-4a) = a^4 - 4a^3 - 9a^2 + 36a...

 

When you do this for the numerator there will be terms that can combine because they have equal powers of 'a'.

 

In general when we have a product of two quadratic polynomials in the same variable:

(Ax^2+Bx+C)(Fx^2+Gx+H)=AFx^4+AGx^3+BFx^3+AHx^2+CFx^2BGx^2+BHx+CGx+CH

 

Note how I've reordered the terms in the expanded sum so that we are ready to group the x^3, x^2, and x terms together to get single coeefficients...as follows:

AGx^3+BFx^3 = (AG+BF)x^3

AHx^2+CFx^2BGx^2 = (AH+CF+BG)x^2

BHx+CGx = (BH+CG)x

 

This gives us the formula for multiplying arbitrary quadratic functions:

(Ax^2+Bx+C)(Fx^2+Gx+H)=AFx^4+(AG+BF)x^3+(AH+CF+BG)x^2+(BH+CG)x+CH

 

Note that the numerator in your problem was written:

(a2-a-12)(a2-4a+3)

where A=1 B=-1 C=-12 F=1 G=-4 H=3

 

Substituting should allow you to complete this problem.