# Literal calculation: answer key to 3rd grade math exercises in PDF.

The answer key to the math exercises in 3ème on literal calculation. Develop or factor a literal expression and solve computational programs.

Exercise 1:

Expand using remarkable identities :

Exercise 2:

We give A = (3x-5) (6-4x)-5(8-6x)

1) Expand and reduce A .

2) Calculate the exact value of A if ; then give the value rounded to the hundredth .

Exercise 3:

Expand and reduce the following expressions:

Exercise 4:

Expand, then reduce, if possible, each expression:

Exercise 5:

Calculate without a calculator and without performing operations:

1. 101²=(100+1)²=100²+2x100x1+1²=10 000+ 200 +1= 10 201

2. 103²=(100+3)²=100²+2x3x100+3²=10 000 + 600 + 9= 10 609

3. 98²=(100-2)²=100²-2x100x2+2²=10 000 – 400 + 4= 9 604

4. 101×99=(100+1)(100-1)=100²-1²= 9 999

Exercise 6:
Expand the following literal expressions:

Exercise 7:

Develop these literal expressions and detail all the steps:

a) (x-1)²= x²-2x+1

b) (x+4)²= x²+8x+16

c) (2x+1)²=4x²+4x+1

d) (7x-1)(7x+1)=49x²-1

e) (4x-1)(3x+7)=12x²+28x-3x-7=12x²+25x-7

f) (-x+1)(3x-2)=-3x²+2x+3x-2=-3x²+5x-2

g) (1/2+x)²=

h) (x-4)²+(x+2)(x+3)=x²-8x+16+x²+3x+2x+6=2x²-3x+22

i) (5x-3)(2x+1)-(x+1)²= ?

Exercise 8:

Expand and reduce the following expressions:

Exercise 9:

Expand with remarkable identities

and reduce the expressions :

Exercise 10:

Factor the following literal expressions:

Exercise 11:

Factor the following expressions:

Exercise 12:

We consider the expression :

1. Factorize D.

2. Expand and reduce D.

3. Calculate D for x = – 1 .

Exercise 13:

We consider the expression :

1. Expand and reduce the expression E.

2. Factorize E.

3. Calculate E for x = – 2.

Exercise 14:

Let the following expression be used:

1. Expand and reduce the expression B.

2. Calculate the expression B for :

a. a=1;

b. a=0,75;

c. a=0 .

Exercise 15:

Factor the following literal expressions:

Exercise 16:

Factor the following literal expressions:

Exercise 17:

Factor the following expressions:

Exercise 18:

Factor the following expressions:

Exercise 19:

Expand the following expressions:

then factor them.

Exercise 20:
1. Factorize :

a. 9-12x+4x²=(3-2x)² .

b. (3-2x)²-4 =(3-2x-2)(3-2x+2)=(1-2x)(5-2x).

2. Deduce a factorization of : E = (9-12x+4x²)-4 =(3-2x)²-4=(1-2x)(5-2x).

Exercise 21:

Consider the expressions E = x² – 5x + 5 and F = (2x – 7)(x – 2) – (x – 3)² .

a) Calculate E and F for x = 4.

b) Develop F. The results obtained in question a) are they surprising?

The previous results are not surprising since E=F for any relative number x.

We want to calculate in column B the values taken by the expression E for the values of x written in column A.

What formula must be entered in cell B2 to perform the desired calculation?
(the formula should be able to be extended to the cells below)

Exercise 22:

Consider the expression .

1) Expand and reduce D.

2) Factorize D.

3) Solve the equation .

Property: A product of factors is zero if and only if at least one of the factors is zero.

4) Calculate the exact value of D when .

Let’s take the expanded and reduced form from question 1.

.

Exercise 23:

1. Factor these expressions :

A=36-25x²=(6+5x)(6-5x)

B=100+60x+9x²=(10+3x)²

C=b²-10b²+25=25-9b²=(5-3b)(5+3b)

E=(2-x)²+(2-x)(9-x)=4-4x+x²+18-2x-9x+x²=2x²-15x+22

2. Expand the following literal expressions:

A=(2x-5)²=4x²-20x+25

B=(5x-3)(5x+3)=25x²-9

C=(-3x+5)²=9x²-30x+25

D=(-6x+9)²=36x²-108x+81

Exercise 24:

1. Expand then reduce A.

2. Factorize A.

3. Solve the equation: (2x – 3)(-x + 2) = 0

Exercise 25:

We give: D = (2x – 3)(5x + 4) + (2x – 3)².

1. Show, by detailing the calculations, that D can be written :

2. Solve the equation: (2x – 3)(7x + 1) = 0.

A product of factors is zero if and only if

at least one of the factors is zero.

Exercise 26:

Let E=(3x+2)²-(3x+2)(x+7) .

a) Develop and reduce E .

E=9x²+12x+4-(3x²+21x+2x+14)=9x²+12x+4-3x²-23x-14= 6x²-11x-10

b) Factorize E .

E=(3x+2)(3x+2)-(3x+2)(x+7)=(3x+2)[3x+2-(x+7)]=(3x+2)(3x+2-x-7)= (3x+2)(2x-5)

c) Calculate E for .

E=(3x+2)(2x-5)

Exercise 27:

Complete using the remarkable identities .

A= (3x+5)²=9x²+30x+25

B= (2x-6)²=4x²-24x+36

C=(2x-4)² = 4x²-16y+16

D= 49a²+70a+25=(7a+5

E = 4x²-1=(2x-1)(2x+1)

Exercise 28:
A = (X + 5) ²
A = X²+2x5xX+5²
A = X²+10X+25

B = (3X – 7) ²
B = (3X)²-2x7x3X+7²
B = 9X²-42X+49

C = (X + 4) (X – 4)
C = X²-4²
C = X²-16

D = (9b + 7) ²
D = (9b)²+2x7x9b+7²
D = 81b²+126b+49

E = (7X + 1) (7X – 1)
E = (7X)²-1²
E = 49X²-1

Exercise 29:
Factor with the help of remarkable identities.

A = X² + 6X + 9
A= X²+2x3xX+3²
A= (X+3)²

B = 9X² – 12X + 4
B= (3X)²-2x3Xx2+2²
B= (3X-2)²

C = y² – 9
C= y²-3²
C= (y-3)(y+3)

D = 16a² – 81
D= (4a)²-9²
D= (4a-9)(4a+9)

E = 49a² +70x +25
E= (7a)²+2x7ax5+5²
E= (7a+5)²

F = 144 – 121a²
F= 12²-(11a)²
F= (12-11a)(12+11a)

G = (2X + 5)² – 9
G= (2X + 5)² – 3²
G= (2X+5+3)(2X+5-3)
G= (2X+8)(2X+2)

H = (2X + 1)² – (3X + 5)²
H= (2X+1+3X+5)(2X+1-3X-5)
H= (5X+6)(-X-4)

Exercise 30:
A = 102²
A= (100+2)²
A= 100²+2x100x2+2²
A= 10 000+400+4
A= 10 404

B = 99×101
B= (100-1)(100+1)
B= 100²-1²
B= 10 000-1
B = 9 999

C = 99²
C= (100-1)²
C= 100²-2x100x1+1²
C= 10 000-200+1
C = 9 801

Exercise 31:
1. Express the area A as a function of x .
A=9×4-0.5xXx2X
A=36-X²
A=6²-X²
A=(6-X)(6+X)

2. Express the area B as a function of x .
B = 8×6-0.5x2Xx8
B = 48-8X
B = 8(6-X)

3. For what value(s) of x are these two areas equal?
A = B
is equivalent to
(6-X)(6+X) = 8(6-X)
(6-X)(6+X)-8(6-X) = 0
(6-X)(6+X-8) = 0
(6-X)(X-2) = 0
It’s a product equation.
A product of factors is zero if and only if at least one of the factors is zero.
6-X=0 or X-2=0
X=6 or X=2

Conclusion:

The areas of figures A and B are equal for X = 2 m or X = 6 m.

Exercise 32:
We give E = (2X+3)² – 16.
1. E = (2X)²+2x2Xx3+3²-16
E = 4X²+12X+9-16
E = 4X²+12X-7

2.
For X = 2 : E = 4×2²+12×2-7=16+24-7 = 33.
For X= 1 : E = 4×1²+12×1-7 = 4+12-7 = 16-7= 9.

3.
E = (2X+3-4)(2X+3+4)
E = (2X-1)(2X+7)

4. E = 4X²+14X-2X+7 = 4X²+12X-7
We find the result of question 1. (and fortunately…..)

Exercise 33:
1. Calculate A and B, giving the result as irreducible fractions.

.

2. Consider the expression :

a. Expand and reduce C .

b. Factorize the expression C .

c. solve the equation: (2x-5)(2-x)=0 .

The solutions are

Exercise 36:

A calculation program is given:

1. Choose a number.

Multiply the sum obtained by the chosen number.

4. Add 4 to this product.

5. Write the result.

1. Write the calculations to verify that if we run this program

with the number – 2 then we get 0.

1) -2

2) -2+4=2

3) 2x(-2)=-4

4)-4+4=0

5) 0

2. Give the result provided by the program when the chosen number is 5.

1)5

2)5+4=9

3)9×5=45

4)45+4=49

5) 49

3. We note x the chosen number

What is the literal expression obtained by performing this program?

1)

2)

3)

4)

Give the result in expanded form.

Exercise 38:

The problem is this. We have an equilateral triangle ABC, a point M, of bucolic mood which walks in the triangle.
We call D, E and F the feet of the perpendiculars at M to the three sides of the triangle.
Using geogebra, we see that MD+ME+MF is constant.

Question : Where must M be placed so that the sum MD+ME+MF is minimal ?

Exercise 39:

Construct a square with an area twice that of the square above.

Solutions:

By calculation:

The area is so the area of the square to be constructed must have the value .

Its side has therefore the length .

so it is enough to take as side the length of the diagonal of the square of departure.

Geometric:

either by cutting or by construction :

Exercise 40:

Jacques had a rectangular pool built.

He tiled the edge of this pool.

The lengths are expressed in meters.

1) Express as a function of the area of the pool surface.

The length of the pool is .

The width of the pool is

2) Express in terms of the area of the tiled surface.

4) Calculate the areas and for x=2 m .

Exercise 42:
1) Solve the inequation: and represent the solutions on a graduated line.

The solution set corresponds to all relative numbers greater than or equal to 4.

2) x being a number greater than or equal to 4,

ABCD is a square whose side measures 2x – 3.

a. Show that the area of the rectangle BCEF is expressed by the formula :

The shaded area is the area of square ABCD minus the area of rectangle AFED.

b. Develop and reduce A.

c. Factor A.

d. Solve the equation: (2x – 3)(x – 4) = 0

Property: A product of factors is zero if and only if at least one of the factors is zero.

or

or

e. For what values of x is the area of the rectangle BCEF zero? Justify.

The area of BCFE is zero for or .

Exercise 43:
The following calculation program is given:

– Choose a number.
– Calculate the square of the result.
– Subtract the square from the starting number.
– Subtract 1.

1. a. Perform this program when the chosen number is 10 and show that you get 20.

b. Perform this program when the chosen number is -3 and show that you get -6.

c. Perform this program when the number chosen is 1.5.

2. What conjecture can be made about the result provided by this calculation program?

The result is double the number of starts

Prove this conjecture.

Let x be the starting number.

The calculation program gives us the following literal expression:

Exercise 44:

Riyanne states:

“For any integer N the expression of is always different from zero”.

This is true because (the square of a number is always positive or zero)

We can even say that for any relative number, this expression is greater than or equal to 140.

Exercise 45:

Show that the area of the ring of center O shown below is equal to

The area of the crown is the area of the large disk minus the area of the small disk.

use the remarkable identity

Exercise 46:

1. Calculate the colored areas of the two figures below as a function of x .

Figure orange:

Figure Green:

2. What do we notice?

These two figures have exactly the same area.

## The answer key to the math exercises on literal calculation in 3rd grade.

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