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Factorisation of the polynomial sqrt(3)x...

Factorisation of the polynomial `sqrt(3)x^(2)+11x+6sqrt(3)`

A

`(sqrt(3)x+2)(x-3sqrt(3))`

B

`(sqrt(3)x+2)(x+3sqrt(3))`

C

`(sqrt(2)x+3)(x+2sqrt(3))`

D

`(sqrt(2)x-2)(x+3sqrt(2))`

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AI Generated Solution

The correct Answer is:
To factor the polynomial \( \sqrt{3}x^2 + 11x + 6\sqrt{3} \), we will use the middle term splitting method. Here’s a step-by-step solution: ### Step 1: Identify the coefficients The given polynomial is \( \sqrt{3}x^2 + 11x + 6\sqrt{3} \). - The coefficient of \( x^2 \) (a) is \( \sqrt{3} \). - The coefficient of \( x \) (b) is \( 11 \). - The constant term (c) is \( 6\sqrt{3} \). ### Step 2: Multiply the first and last coefficients We need to multiply the coefficient of \( x^2 \) and the constant term: \[ a \cdot c = \sqrt{3} \cdot 6\sqrt{3} = 6 \cdot 3 = 18 \] ### Step 3: Find two numbers that multiply to \( 18 \) and add up to \( 11 \) We need to find two numbers that multiply to \( 18 \) (from step 2) and add up to \( 11 \) (the coefficient of \( x \)): - The numbers \( 9 \) and \( 2 \) satisfy this condition because: \[ 9 \cdot 2 = 18 \quad \text{and} \quad 9 + 2 = 11 \] ### Step 4: Rewrite the middle term Now, we can rewrite the polynomial by splitting the middle term using the two numbers we found: \[ \sqrt{3}x^2 + 9x + 2x + 6\sqrt{3} \] ### Step 5: Group the terms Next, we group the terms: \[ (\sqrt{3}x^2 + 9x) + (2x + 6\sqrt{3}) \] ### Step 6: Factor out the common terms in each group Now, we factor out the common factors from each group: 1. From the first group \( \sqrt{3}x^2 + 9x \), we can factor out \( 3x \): \[ 3x(\frac{\sqrt{3}}{3}x + 3) = 3x(\sqrt{3}x + 3) \] 2. From the second group \( 2x + 6\sqrt{3} \), we can factor out \( 2 \): \[ 2(x + 3\sqrt{3}) \] ### Step 7: Combine the factored groups Now, we can combine the factored groups: \[ 3x(\sqrt{3}x + 3) + 2(x + 3\sqrt{3}) \] ### Step 8: Factor out the common binomial Notice that both terms contain the common binomial \( x + 3\sqrt{3} \): \[ (x + 3\sqrt{3})(\sqrt{3}x + 2) \] ### Final Answer Thus, the factorization of the polynomial \( \sqrt{3}x^2 + 11x + 6\sqrt{3} \) is: \[ (\sqrt{3}x + 2)(x + 3\sqrt{3}) \]
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MTG IIT JEE FOUNDATION-POLYNOMIALS-EXERCISE (Multiple choice Question (Level-1))
  1. Factors of (a+b)^(3)-(a-b)^(3) are

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  2. The common quantity that must be added to each term of a^(2):b^(2) to ...

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  3. One of the dimensions of the cuboid whose volume is 16x^(2)-26x+10 is

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  4. find the value of x+y+z if x^(2)+y^(2)+z^(2)=18 and xy+yz+zx=9 .

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  5. Find the remainder when the polynomial f(x)=x^(3)-3x^(2)+4x+50 is div...

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  6. The value of a for which (x+a) is a factor of the polynomial x^(3)+ax^...

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  7. Factorisation of the polynomial sqrt(3)x^(2)+11x+6sqrt(3)

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  8. If x=-2 and x^(2)+y^(2)+2xy=0 , then find y .

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  9. If x+(1)/(x)=5 , then find the value of x^(2)+(1)/(x^(2)) .

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  10. Find the value of x^(3)-8y^(3)-36xy-216 , when x=2y+6 .

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  11. Simplify : (x^(3)-4-x+4x^(2))/(x^(2)+3x-4)

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  12. Which of the following is true if (x+1) and (x+2) are factors of p(x)=...

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  13. Factorise : 6x^(3)-5x^(2)-13x+12

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  14. If p(x)=x^(3)-3x^(2)-2x+4 , then find the value of [p(2)+p(-2)-p(0)] .

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  15. If p=2-a , then a^(3)+6ap+p^(3)-8 =

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  16. The polynomial p(x)=x^(4)-2x^(3)+3x^(2)-ax+3a when divided by (x+1) le...

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  17. The values of a and b so that the polynomial x^(3)-ax^(2)-13x+b has (x...

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  18. The product (a+b)(a-b)(a^2-a b+b^2)(a^2+a b+b^2) is equal to: a^6+b^6 ...

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  19. Posible factors of x^(4)+x^(3)-7x^(2)-x+6 are

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  20. If a+b+c=0 , then (a^2)/(b c)+(b^2)/(c a)+(c^2)/(a b)=\ 0 (b) 1 (...

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