
The HCF of $({{x}^{2}}-9),\,({{x}^{3}}-27)\:\text{and}\:({{x}^{2}}-8x+15)$ is
\[\text{A}.\,\,\,x-3\]
$\text{B}.\,\,\,x+3$
$\text{C}.\,\,\,(x-3)({{x}^{2}}-3x+9)$
$\text{D}.\,\,\,(x-3)(x+3)(x+5)({{x}^{2}}+3x+9)$
Answer
485.4k+ views
Hint: find the factor or rather factorize them. The four methods of solving a quadratic equation are factoring, using the square roots, completing the square and the quadratic formula.
Formula and
\[\begin{align}
& {{a}^{2}}-{{b}^{2}}=(a-b)(a+b) \\
& {{a}^{3}}-{{b}^{3}}=(a-b)({{a}^{2}}-ab+{{b}^{2}}) \\
& {{a}^{3}}+{{b}^{3}}=(a-b)({{a}^{2}}-ab+{{b}^{2}}) \\
\end{align}\]
Also we can add ${{a}^{3}}-{{b}^{3}}={{(a-b)}^{3}}+3{{a}^{2}}b-3a{{b}^{2}}$
Complete step by step solution: In mathematics, factorization or factoring consists of writing a number or another mathematical object as a product of several factors, usually smaller or simpler objects of the same kind. For example, 3 × 5 is a factorization of the integer 15, and is a factorization of the polynomial x² – 4.
$({{x}^{2}}-9);\,({{x}^{3}}-27)\,.\,\text{and}\,({{x}^{2}}-8x+15).({{x}^{2}}-9)$
So, this is the given part to which we have to find the Hcf
So, using the formula $[{{a}^{2}}-{{b}^{2}}=(a-b)(a+b)]$
We have
\[\begin{align}
& {{x}^{2}}-9 \\
& \Rightarrow {{x}^{2}}-{{3}^{2}}=(x-3)(x+3) \\
\end{align}\]
Also, using the formula ${{a}^{3}}-{{b}^{3}}=(a-b)({{a}^{2}}-ab+{{b}^{2}})$
We have
\[\begin{align}
& {{x}^{3}}-27 \\
& \Rightarrow {{x}^{3}}-33=(x-3)({{x}^{2}}-2x+3) \\
\end{align}\]
Now, if we factorize ${{x}^{2}}-8x+15$
We get
\[\begin{align}
& {{x}^{2}}-8x+15={{x}^{2}}-5x-3x+15 \\
& =x(x-5)-3(x-5) \\
& =(x-3)(x-5)
\end{align}\]
Therefore, we can see $(x-3)$is the only common factor in $({{x}^{2}}-9),\,({{x}^{2}}-27)({{x}^{2}}-8x+15)$
Hence the HCF for this question is $(x-3)$
Therefore option (A) is the correct answer.
Note: In this type of question, factorization is must know, find factors then see for the common factors and as HCF suggest the common factor is the HCF of the question. The highest common factor is found by multiplying all the factors which appear in both lists: So the HCF of 60 and 72 is 2 × 2 × 3 which is 12. The lowest common multiple is found by multiplying all the factors which appear in either list: So the LCM of 60 and 72 is 2 × 2 × 2 × 3 × 3 × 5 which is 360.
Formula and
\[\begin{align}
& {{a}^{2}}-{{b}^{2}}=(a-b)(a+b) \\
& {{a}^{3}}-{{b}^{3}}=(a-b)({{a}^{2}}-ab+{{b}^{2}}) \\
& {{a}^{3}}+{{b}^{3}}=(a-b)({{a}^{2}}-ab+{{b}^{2}}) \\
\end{align}\]
Also we can add ${{a}^{3}}-{{b}^{3}}={{(a-b)}^{3}}+3{{a}^{2}}b-3a{{b}^{2}}$
Complete step by step solution: In mathematics, factorization or factoring consists of writing a number or another mathematical object as a product of several factors, usually smaller or simpler objects of the same kind. For example, 3 × 5 is a factorization of the integer 15, and is a factorization of the polynomial x² – 4.
$({{x}^{2}}-9);\,({{x}^{3}}-27)\,.\,\text{and}\,({{x}^{2}}-8x+15).({{x}^{2}}-9)$
So, this is the given part to which we have to find the Hcf
So, using the formula $[{{a}^{2}}-{{b}^{2}}=(a-b)(a+b)]$
We have
\[\begin{align}
& {{x}^{2}}-9 \\
& \Rightarrow {{x}^{2}}-{{3}^{2}}=(x-3)(x+3) \\
\end{align}\]
Also, using the formula ${{a}^{3}}-{{b}^{3}}=(a-b)({{a}^{2}}-ab+{{b}^{2}})$
We have
\[\begin{align}
& {{x}^{3}}-27 \\
& \Rightarrow {{x}^{3}}-33=(x-3)({{x}^{2}}-2x+3) \\
\end{align}\]
Now, if we factorize ${{x}^{2}}-8x+15$
We get
\[\begin{align}
& {{x}^{2}}-8x+15={{x}^{2}}-5x-3x+15 \\
& =x(x-5)-3(x-5) \\
& =(x-3)(x-5)
\end{align}\]
Therefore, we can see $(x-3)$is the only common factor in $({{x}^{2}}-9),\,({{x}^{2}}-27)({{x}^{2}}-8x+15)$
Hence the HCF for this question is $(x-3)$
Therefore option (A) is the correct answer.
Note: In this type of question, factorization is must know, find factors then see for the common factors and as HCF suggest the common factor is the HCF of the question. The highest common factor is found by multiplying all the factors which appear in both lists: So the HCF of 60 and 72 is 2 × 2 × 3 which is 12. The lowest common multiple is found by multiplying all the factors which appear in either list: So the LCM of 60 and 72 is 2 × 2 × 2 × 3 × 3 × 5 which is 360.
Recently Updated Pages
What percentage of the area in India is covered by class 10 social science CBSE

The area of a 6m wide road outside a garden in all class 10 maths CBSE

What is the electric flux through a cube of side 1 class 10 physics CBSE

If one root of x2 x k 0 maybe the square of the other class 10 maths CBSE

The radius and height of a cylinder are in the ratio class 10 maths CBSE

An almirah is sold for 5400 Rs after allowing a discount class 10 maths CBSE

Trending doubts
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Why is there a time difference of about 5 hours between class 10 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE

Write an application to the principal requesting five class 10 english CBSE
