please answer me my question don't fake answer
solve karna hai tick mark
Answers
1. The factorisation of 4x² + 8x + 3 is
Answer:
First, we need to split the middle term. To do so we will follow the below steps.
Sum: Middle term with variable with the power '1'
Product: Product of the constants of the two other terms.
So,
Sum: 8
Product: 4 × 3 = 12
Now we need to find the pair of two number which when multiplied gives us the product number and when added/subtracted gives the sum number. To find this number we will first list out all numbers when multiplied gives 12.
1 & 12
2 & 6
3 & 4
So here the possible pair is 2 & 6 that meets our requirements. So when we spilt the middle term we will get 2x and 6x.
→ 4x² + 8x + 3
→ 4x² + 2x + 6x + 3
Common Factors
4x² = 2 × 2 × x × x
2x = 2 × x
Common factor = 2x
→ 4x² + 2x + 6x + 3
→ 2x (2x + 1) + 6x + 3
Common Factors
6x = 2 × 3 × x
3 = 1 × 3
Common Factor = 3
→ 2x (2x + 1) + 6x + 3
→ 2x (2x + 1) + 3 (2x + 1)
→ (2x + 3)(2x + 1)
∴ The factorisation of 4x² + 8x + 3 is b) (2x + 1)(2x + 3)
________________________________
2. Which of the following is a factor of (x + y)³ - (x³ + y³)
Answer:
To find the factor we must simplify it first.
Let's simplify using identities.
Identity -: (a + b)³ = a³ + b³ + 3ab (a + b)
→ x³ + y³ + 3xy (x + y) - (x³ + y³)
→ x³ + y³ + 3xy (x + y) - x³ - y³
→ x³ - x³ + y³ - y³ + 3xy (x + y)
→ 3xy (x + y)
∴ The factor of (x + y)³ - (x³ + y³) is d) 3xy
________________________________
3. One of the factors of (25x² - 1) + (1 + 5x)² is
Answer:
To find the factor we must simplify it first.
Let's simplify using identities.
→ (25x² - 1) + (1 + 5x)²
→ 25x² - 1 + (1 + 5x)²
Identity → (a + b)² = a² + 2ab + b²
→ 25x² - 1 + (1 + 5x)²
→ 25x² - 1 + (1)² + 2(1)(5x) + (5x)²
→ 25x² - 1 + 1 + 10x + 25x²
→ 25x² + 25x² - 1 + 1 + 10x
→ 50x² + 10x
Now let's factorize it.
Common Factors
50x² = 2 × 5 × 5 × x × x
10x = 2 × 5 × x
Common Factor = 10x
→ 50x² + 10x
→ 10x (5x + 1)
∴ One of the factors of (25x² - 1) + (1 + 5x)² is d) 10x
________________________________
4. If x - 3 is a factor of 2x² - kx + 6, then the value of k is
Answer
x - 3 = 0
x = 3
Let's substitue the value of 'x' in the equation and find the value of 'k'
2(3)² - k(3) + 6 = 0
2(9) - 3k + 6 = 0
18 - 3k + 6 = 0
18 + 6 - 3k = 0 [Combine Like terms]
24 - 3k = 0
24 - 3k - 24 = 0 - 24 [Subtract 24 from both sides of the equation]
-3k = -24
3k = 24
3k/3 = 24/3 [Divide 3 from both sides of the equation]
k = 8
∴ If x - 3 is a factor of 2x² - kx + 6, then the value of k is d) 8
________________________________
5. One of the factors of (x - 1)³ - (x² - 1) is
Answer
First we'll simplify the expression.
→ (x - 1)³ - (x² - 1)
→ (x- 1)³ + -1 (x² - 1)
→ (x - 1)³ + -1x² + (-1)(-1)
→ (x - 1)³ + -x² + 1
→ x³ + -3x² + 3x + -1 + -x² + 1
→ (x³) + (-3x² + -x²) + (3x) + (-1 + 1)
→ x³ + -4x² + 3x
→ x³ - 4x² + 3x
Common Factors
x³ = x × x × x
4x² = 2 × 2 × x × x
3x = 3 × x
Common Factor = x
→ x (x² - 4x + 3)
Split the middle term.
Sum -: 4
Product -: 3
→ x (x² - 1x - 3x + 3)
→ x [x² - 1x - 3x + 3]
Common Factors
x² = x × x
1x = 1 × x
Common Factor = x
→ x [x² - 1x - 3x + 3]
→ x [x (x - 1) - 3x + 3]
Common Factors
3x = 3 × x
3 = 1 × 3
Common Factor = 3
→ x [x (x - 1) - 3x + 3]
→ x [x (x - 1) - 3 (x + 1)]
→ x [(x - 1)(x - 3)]
→ x (x - 1)(x - 3)
∴ One of the factors of (x - 1)³ - (x² - 1) is b) x - 3
________________________________
Hope it helps you!
Step-by-step explanation:
Question-1:
The factorisation of 4x²+8x+3 is:
On splitting the middle terms,
=> 4x²+2x+6x+3
=> 2x(2x+1)+3(2x+1)
=> (2x+3)(2x+1)
Option B is the right answer.
Question-2:
(x+y)³-(x³+y³)
=> x³+y³+3xy(x+y)-x³-y³
x³ and y³ gets cancelled,
=> 3xy(x+y)
Hence, 3xy is the factor of the given question.
Option-D
Question-3
One of the factors of (25x²-1)+(1+5x)² is:
=> 25x²-1+1+25x²+10x
=> 50x²+10x
=> 10x[5x+1]
Hence, the factor is 10x
Option D
Question-4:
=> x-3 = 0
=> x = 3
Substitute the values:
=> f(3) = 2(3²)-k(3)+6 = 0
=> 18-3k+6 = 0
=> 24-3k = 0
=> -3k = -24
=> k = 8
Option D
Question-5
=> (x-1)³-(x²-1)
=> x³-1-3x(x-1)-x²+1
=> x³-3x²+3x-x²
=> x³-4x²+3x
=> x(x²-4x+3)
=> x(x²-3x-x+3)
=> x[x(x-3)-1(x-3)]
=> x(x-1)(x-3)
Option B