# Simplify 7x^{2}(3x – 9) + 3 and find its values for x = 4 and x = 6

An algebraic expression, which is also known as a variable expression is an equation composed of variable terms formed from the combination of constants and variables. These components are joined together using operations, like addition, subtraction, multiplication, or division. The constants accompanied by the variable in each term are referred to as the coefficient.

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**Solve the equation 7x**^{2}(3x – 9) + 3 for x = 4 and x = 6

^{2}(3x – 9) + 3 for x = 4 and x = 6

**Solution: **

⇒ 7x

^{2}(3x – 9) + 3Find the solution for 7x

^{2}(3x – 9)By using distributive law which states that;

a(b – c) = ab – ac

So according to the law

⇒ (7x

^{2}× 3x) – (7x^{2}× 9)⇒ 21x

^{3}– 63x^{2}Therefore,

⇒ 7x

^{2}(3x – 9) + 3 = 21x^{3}– 63x^{2}+ 3Now we have to solve for the equation

21x

^{3}– 63x^{2}+ 3Further,

For x = 4,

⇒ 21x

^{3}– 63x^{2}+ 3⇒ 21 × 4

^{3}– 63 × 4^{2}+ 3⇒ 1344 – 1008 + 3

⇒ 336 + 3

⇒ 339

Further,

Now for x = 6,

⇒ 21x

^{3}– 63x^{2}⇒ 21 × 6

^{3}– 63 × 6^{2}+ 3⇒ 2268 + 3

⇒ 2271

Therefore,

The algebraic expression ⇒ 7x

^{2}(3x – 9) + 3For the value of x = 4 is 339

For the value of x = 6 is 2271

### Sample Questions

**Question 1. By applying suitable algebraic identity, Find 1050 ^{2}**

**Solution:**

By applying the algebraic identity in the question: (a + b)² = a² + 2ab + b²

Thus,

1050 = 1000 + 50

Therefore,

1050

^{2}= (1000 + 50)^{2}Here,

a = 1000

b = 50

(1000 + 50)

^{2}= (1000)² + 2 × 1000 × 50 + (50)²= 1000000 + 100000 + 2500

Therefore,

1050

^{2}= 1102500.

**Question 2. Simplify 82 + 2×(5x – 7). For the values of x = 2 and x = -2?**

**Solution:**

Here we have,

82 + 2 × (5x – 7)

For x = 2

Substitute value of x = 2 in the equation

= 82 + 2 × (5x – 7)

= 82 + (2 × 5x – 2 × 7)

= 82 + (10x – 14)

= 82 + 10x – 14

= 82 – 14 + 10 × 2

= 82 – 14 + 20

= 88

For x = -2

Substitute value of x = -2 in the equation

= 82 + 2 × (5x – 7)

= 82 + (2 × 5x – 2 × 7)

= 82 + (10x – 14)

= 82 + 10x – 14

= 82 – 14 + 10 × (-2)

= 82 – 14 – 20

= 48

**Question 3. Simplify 24 × 7 + x(365 – 65). For the value of x = 1 and x = -1**

**Solution:**

Here we have

24 × 7 + x(365 – 65)

For x = 1

Substitute value of x = 1 in the equation

= 24 × 7 + x(365 – 65)

= 168 + x(365 – 65)

= 168 + 365x – 65x

= 168 + 300x

= 168 + 300 × 1

= 168 + 300

= 468

For x = -1

Substitute value of x = -1 in the equation

= 24 × 7 + x(365 – 65)

= 168 + x(365 – 65)

= 168 + 365x – 65x

= 168 + 300x

= 168 + 300 × (-1)

= 168 – 300

= -132

**Question 4. Subtract the polynomials.**

**(6x + 3) from (-8x + 6)**

**And simplify for x = 4**

**Solution:**

(6x + 3) from (-8x + 6)

= (-8x + 6) – (6x + 3)

= -8x + 6 – 6x – 3

= -8x -6x + 6 – 3

= -14x + 3

For x = 4

Substitute value of x = 4 in the equation

= -14 × 4 + 3

= -56 + 3

= -53

**Question 5. Solve the equation 5x ^{2}(6x – 7) + 5 for x = 2 and x = 4**

**Solution:**

5x

^{2}(6x – 7) + 5Find the solution for

5x

^{2}(6x – 7)By using distributive law which states that;

a(b – c) = ab – ac

So according to the law

⇒ (5x

^{2}× 6x) – (5x^{2}× 7)⇒ 30x

^{3}– 35x^{2}Therefore,

⇒ 5x

^{2}(6x – 7) + 5 = 30x^{3}– 35x^{2}+ 5Now we have to solve for the equation

30x

^{3}– 35x^{2}+ 5Further,

For x = 4,

⇒ 30x

^{3}– 35x^{2}+ 5⇒ 30 × 4

^{3}– 35 × 4^{2}+ 5⇒ 1920 – 560 + 5

⇒ 1360 + 5

⇒ 1365

Further,

Now for x = 6,

⇒ 30x

^{3}– 35x^{2}+ 5⇒ 30 × 6

^{3}– 35 × 6^{2}+ 5⇒ 6480 – 1260 + 5

⇒ 5220 + 5

⇒ 5225

Therefore,

The algebraic expression ⇒ 5x

^{2}(6x – 7) + 5For the value of x = 4 is 1365

For the value of x = 6 is 5225