Multiplying Binomials (FOIL) Test Prep AK

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1. D

$\eqalign{(4x+5)(3x-7)\\12x^2-28x+15x-35\\12x^2-13x-35}$

 

2. C

$\eqalign{3(2x-1)(2x+2)\\3(4x^2+4x-2x-2)\\3(4x^2+2x-2)\\12x^2+6x-6}$

3. 4

To make the fraction undefined, the denominator must be 0.

$\eqalign{(x-3)^2 - 2(x-3)+1=&0\\x^2-6x+9-2x+6+1=&0\\x^2-8x+16=&0\\(x-4)(x-4)=&0\\x=&4}$

 

4. 7

To make the fraction undefined, the denominator must be 0.

$\eqalign{(x+2)^2 - 6(3x-7)+3=&0\\x^2+4x+4-18x+42+3=&0\\x^2-14x+49=&0\\(x-7)(x-7)=&0\\x=&7}$

 

5. D

First, FOIL the binomals:

$\eqalign{(ax+1)(bx+9)&=6x^2+cx+9\\abx^2+9ax+bx+9=6x^2+9ax+bx+9}$

By matching up the 2 equations, you can see that $ab=6$ and the problem says that $a+b=5$ so $a$ and $b$ must equal 2 and 3. 

$c$ is the combination of the two middle terms from FOIL.  $c=9a+b$ so the two possible answers will be the sum of $9a+b$.  Try inserting each possible answer:

$9(3)+2=29$

$9(2)+3=21$