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PSAT Math : How to use FOIL in the distributive property

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Example Questions

Example Question #1 : Foil

$(\sqrt5+\sqrt3)^2=$

Possible Answers:

$8+\sqrt{15}$

$2+\sqrt{15}$

$8+2\sqrt{15 }$

$\sqrt{15}$

$2+2\sqrt{15}$

Correct answer:

$8+2\sqrt{15 }$

Explanation:

$(\sqrt5+\sqrt3)^2=(\sqrt5)^2+2\sqrt5\sqrt3+(\sqrt3)^2=5+2\sqrt{15}+3=8+2\sqrt{15}$

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Example Question #1 : How To Use Foil In The Distributive Property

If (3x - 4)(3x + 4) = 2 , what is the value of 9x^2 ?

Possible Answers:

Correct answer:

Explanation:

Remember that (a – b )(a + b ) = a ²– b ².

We can therefore rewrite (3x – 4)(3x + 4) = 2 as (3x )²– (4)²= 2.

Simplify to find 9x²– 16 = 2.

Adding 16 to each side gives us 9x²= 18.

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Example Question #3 : Foil

If g(x)=2x^2-2 and h(x)=x+4 , then which of the following is equivalent to g(h(x))-h(g(x)) ?

Possible Answers:

16x+28

2x^3+8x^2-2x-8

16x

-16x-28

Correct answer:

16x+28

Explanation:

We are asked to find the difference between g(h(x)) and h(g(x)), where g(x) = 2x² – 2 and h(x) = x + 4. Let's find expressions for both.

g(h(x)) = g(x + 4) = 2(x + 4)² – 2

g(h(x)) = 2(x + 4)(x + 4) – 2

In order to find (x+4)(x+4) we can use the FOIL method.

(x + 4)(x + 4) = x² + 4x + 4x + 16

g(h(x)) = 2(x² + 4x + 4x + 16) – 2

g(h(x)) = 2(x² + 8x + 16) – 2

Distribute and simplify.

g(h(x)) = 2x² + 16x + 32 – 2

g(h(x)) = 2x² + 16x + 30

Now, we need to find h(g(x)).

h(g(x)) = h(2x² – 2) = 2x² – 2 + 4

h(g(x)) = 2x² + 2

Finally, we can find g(h(x)) – h(g(x)).

g(h(x)) – h(g(x)) = 2x² + 16x + 30 – (2x² + 2)

= 2x² + 16x + 30 – 2x² – 2

= 16x + 28

The answer is 16x + 28.

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Example Question #4 : Foil

The sum of two numbers is . The product of the same two numbers is . If the two numbers are each increased by one, the new product is . Find q-p in terms of .

Possible Answers:

2s+1

s+1

s^2

s+2

s-1

Correct answer:

s+1

Explanation:

Let the two numbers be x and y.

x + y = s

xy = p

(x + 1)(y + 1) = q

Expand the last equation:

xy + x + y + 1 = q

Note that both of the first two equations can be substituted into this new equation:

p + s + 1 = q

Solve this equation for q – p by subtracting p from both sides:

s + 1 = q – p

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Example Question #2 : How To Use Foil In The Distributive Property

Expand the expression:

$\dpi{100} \small (x^{3}-4x)(6 + 12x^{2})$

Possible Answers:

$\dpi{100} \small 12x^{5}-42x^{3}-24x$

$\dpi{100} \small 22x^{2}$

$\dpi{100} \small 6x^{3} + 12x^{2}-24x-48$

$\dpi{100} \small 6x^{3} + 12x^{5}-24x-48x^{3}$

$\dpi{100} \small 42x^{3}+12x^{5}-24x$

Correct answer:

$\dpi{100} \small 12x^{5}-42x^{3}-24x$

Explanation:

When using FOIL, multiply the first, outside, inside, then last expressions; then combine like terms.

$\dpi{100} \small (x^{3}-4x)(6 + 12x^{2})$

$\dpi{100} \small 6x^{3}+12x^{5}-24x-48x^{3}$

$\dpi{100} \small -42x^{3}+12x^{5}-24x$

$\dpi{100} \small 12x^{5}-42x^{3}-24x$

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Example Question #2 : How To Use Foil In The Distributive Property

Expand the following expression:

$(4x+2)(x^2-2)$

Possible Answers:

$4x^3+4x-4$

$4x^3+2x^2+8x+4$

$4x^3+2x^2-8x-4$

$x^3+2x^2-8x-4$

$4x^3-4$

Correct answer:

$4x^3+2x^2-8x-4$

Explanation:

$(4x+2)(x^2-2)=(4x\times x^2)+(4x\times -2)+(2\times x^2) +(2\times -2)$

Which becomes

$4x^3-8x+2x^2-4$

Or, written better

$4x^3+2x^2-8x-4$

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Example Question #7 : Foil

Which of the following is equal to the expression (3x - 1)(2x + 4) ?

Possible Answers:

{5x+3}

6x^2+10x-4

6x^2+2x+10

6x^2-4

6x^2-10

Correct answer:

6x^2+10x-4

Explanation:

Multiply using FOIL:

First = 3x(2x) = 6x²

Outter = 3x(4) = 12x

Inner = -1(2x) = -2x

Last = -1(4) = -4

Combine and simplify:

6x² + 12x - 2x - 4 = 6x² +10x - 4

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Example Question #8 : Foil

Simplify the expression.

(2x^2-3x)(2y+a)

Possible Answers:

2x^2y - 2ax^2 - 3xy - 3ax

None of the other answers

-6x^3 + 2ay

4x^2y + 2ax^2 - 6xy - 3ax

-2axy

Correct answer:

4x^2y + 2ax^2 - 6xy - 3ax

Explanation:

Solve by applying FOIL:

First: 2x² * 2y = 4x²y

Outer: 2x² * a = 2ax²

Inner: –3x * 2y = –6xy

Last: –3x * a = –3ax

Add them together: 4x²y + 2ax² – 6xy – 3ax

There are no common terms, so we are done.

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Example Question #3 : How To Use Foil In The Distributive Property

(6x - 2) (3x + 5) = ax^2 + dx + k

Given the equation above, what is the value of a-d+k ?

Possible Answers:

Correct answer:

Explanation:

(6x - 2) (3x + 5) = ax^2 + dx + k

Use FOIL to expand the left side of the equation.

(6x-2)(3x+5)=18x^2+30x-6x-10=18x^2+24x-10

18x^2+24x-10=ax^2+dx+k

From this equation, we can solve for , , and .

$18x^2=ax^2\rightarrow a=18$

$24x=dx\rightarrow d=24$

-10=k

Plug these values into a-d+k to solve.

a-d+k=(18)-(24)+(-10)=-16

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Example Question #1 : Distributive Property

Expand and simplify the expression.

4(x-5)(2x+10)

Possible Answers:

$8x^{2}-200$

$8x^{2}-80x-200$

$8x^{2}-200x$

$8x^{2}-40x+200$

8x^2-10x+10

Correct answer:

$8x^{2}-200$

Explanation:

4(x-5)(2x+10)

We can solve by FOIL, then distribute the $\small 4$ . Since all terms are being multiplied, you will get the same answer if you distribute the $\small 4$ before using FOIL.

First: x*2x=2x^2

Inside: -5*2x=-10x

Outside: x*10=10x

Last: -5*10=-50

Sum all of the terms and simplify. Do not forget the $\small 4$ in front of the quadratic!

4(2x^2-10x+10x-50)

4(2x^2-50)

Finally, distribute the $\small 4$ .

8x^2-200

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PSAT Math : How to use FOIL in the distributive property

Study concepts, example questions & explanations for PSAT Math

All PSAT Math Resources

Example Questions

Example Question #1 : Foil

Example Question #1 : How To Use Foil In The Distributive Property

Example Question #3 : Foil

Example Question #4 : Foil

Example Question #2 : How To Use Foil In The Distributive Property

Example Question #2 : How To Use Foil In The Distributive Property

Example Question #7 : Foil

Example Question #8 : Foil

Example Question #3 : How To Use Foil In The Distributive Property

Example Question #1 : Distributive Property

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