How to use FOIL in the distributive property - PSAT Math

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Question

Given the equation above, what is the value of ?

Answer

Use FOIL to expand the left side of the equation.

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

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

$24x=dx\rightarrow d=24$

Plug these values into to solve.

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Question

Which of the following is equal to the expression ?

Answer

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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Question

If , what is the value of ?

Answer

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

We can therefore rewrite (3_x –_ 4)(3_x_ + 4) = 2 as (3_x_ )2 – (4)2 = 2.

Simplify to find 9_x_2 – 16 = 2.

Adding 16 to each side gives us 9_x_2 = 18.

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Question

If and , then which of the following is equivalent to ?

Answer

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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Question

Simplify the expression.

Answer

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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Question

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 in terms of ___.

Answer

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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Question

Expand the expression:

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

Answer

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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Question

Expand the following expression:

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

Answer

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

Which becomes

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

Or, written better

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

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Question

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

Answer

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

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Question

Expand and simplify the expression.

Answer

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:

Inside:

Outside:

Last:

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

Finally, distribute the $\small 4$ .

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Question

If , , and $h(x) = f(x) \times g(x)$ , then

Answer

To find what equals, you must know how to multiply times , or, you must know how to multiply binomials. The best way to multiply monomials is the FOIL (first, outside, inside, last) method, as shown below:

$h(x) = f(x)\times g(x) = (3x-2)\times (2x+3)$

Multiply the First terms

$3x\cdot 2x= 6x^{2}$

Multiply the Outside terms:

$3x\cdot3 = 9x$

Multiply the Inside terms:

$-2\cdot2x = -4x$

Note: this step yields a negative number because the product of the two terms is negative.

Multiply the Last terms:

$-2\cdot3=-6$

Note: this step yields a negative number too!

Putting the results together, you get:

$(3x-2)\times (2x+3) = 6x^{2} +9x-4x -6$

Finally, combine like terms, and you get:

$h(x) = 6x^{2} +5x-6$

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Question

Given the equation above, what is the value of ?

Answer

Use FOIL to expand the left side of the equation.

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

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

$24x=dx\rightarrow d=24$

Plug these values into to solve.

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Question

Which of the following is equal to the expression ?

Answer

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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Question

If , what is the value of ?

Answer

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

We can therefore rewrite (3_x –_ 4)(3_x_ + 4) = 2 as (3_x_ )2 – (4)2 = 2.

Simplify to find 9_x_2 – 16 = 2.

Adding 16 to each side gives us 9_x_2 = 18.

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Question

If and , then which of the following is equivalent to ?

Answer

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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Question

Simplify the expression.

Answer

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.

Compare your answer with the correct one above

Question

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 in terms of ___.

Answer

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

Compare your answer with the correct one above

Question

Expand the expression:

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

Answer

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

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

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

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

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

Compare your answer with the correct one above

Question

Expand the following expression:

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

Answer

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

Which becomes

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

Or, written better

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

Compare your answer with the correct one above

Question

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

Answer

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

Compare your answer with the correct one above

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