Question

Factorise 2x³ - 13x² + 20x - 7

Answer 1

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✤ Required Answer:

✒ To Factorise:-

• 2x³ - 13x² + 20x - 7

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✤ How to solve?

The cubic polynomials ar generally factorised by assuming the first factor, and then getting the other factors. One of the easiest way to assume is by Factor theoram, So Let's see the statement of Factor's theoram:

• When there is any polynomial p(x) of degree > or = to 1 and a be any real number such that p(a) = 0, Then (x - a) is a factor of p(x). And, Converse !!

☃️ So, By using this let's solve this question....

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✤ Solution:

Let p(x) = 2x³ - 13x² + 20x - 7 be the given polynomial. The constant term is -7 are ± 1, ± 7. But, the factor of coefficient of x³ is 2.

So,

• Possible rational roots of p(x) = ± 1, ± 7, ± 1/2 and ± 7/2

Putting x = 1 in p(x), We get

➝ p(1) = 2(1)³ - 13(1)² + 20(1) - 7

➝ p(1) = 2 - 13 + 20 - 7

➝ p(1) = 2 - 13 + 13

➝ p(1) = 2 not equals to 0

So, Let's try putting x = 1/2 in p(x),

➝ 2(1/2)³ - 13(1/2)² + 20(1/2) - 7

➝ 2 × 1/8 - 13/4 + 10 ‐ 7

➝ 1/4 - 13/4 + 3

➝ -12/4 + 3

➝ 0

|| (x - 1/2) is a factor of p(x) by factor's theoram ||

➝ (x - 1/2) = (2x - 1/2)

➝ So, (2x - 1) is also the factor.

Now arranging to get (2x - 1) in p(x),

➝ 2x³ - 13x² + 20x - 7

➝ 2x³ - x² - 12x² + 6x + 14x - 7

➝ x²(2x - 1) - 6x(2x - 1) + 7(2x - 1)

➝ (2x - 1)(x² - 6x + 7)

Further factorising (x² - 6x + 7),

➝ x² - 6x + 7

By sreedhar acharya's formula,

➝ $\large{ \sf{ \dfrac{ - b \pm \sqrt{b {}^{2} - 4ac } }{2a} }}$

➝ $\large{ \sf{ \dfrac{ 6 \pm \sqrt{ {6}^{2} - 4 \times 1 \times 7 } }{2 \times 1} }}$

➝ $\large{ \sf{ \dfrac{6 \pm \sqrt{36 - 28} }{2} }}$

➝ $\large{ \sf{ \dfrac{ - 6 \pm2 \sqrt{2} }{2} }}$

➝ $\large{ \sf{ 3 + \sqrt{2}}} \: \: \& \: \: 3 - \sqrt{2}$

So,

• x = 3 + √2
• x = 3 - √2

Then,

➝ 2x³ - 13x² + 20x - 7

➝ (2x - 1)(x² - 6x + 7)

➝ (2x - 1)[(x - (3 + √2)][x - (3 - √2)]

➝ (2x - 1)(x - 3 - √2)(x - 3 + √2)

☀️ Hence, factorised !!

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Answer 2

✤ Required Answer:

✒ To Factorise:-

2x³ - 13x² + 20x - 7

━━━━━━━━━━━━━━━━━━━━

✤ How to solve?

The cubic polynomials ar generally factorised by assuming the first factor, and then getting the other factors. One of the easiest way to assume is by Factor theoram, So Let's see the statement of Factor's theoram:

When there is any polynomial p(x) of degree > or = to 1 and a be any real number such that p(a) = 0, Then (x - a) is a factor of p(x). And, Converse !!

☃️ So, By using this let's solve this question......

✤ Solution:

Let p(x) = 2x³ - 13x² + 20x - 7 be the given polynomial. The constant term is -7 are ± 1, ± 7. But, the factor of coefficient of x³ is 2.

So,

Possible rational roots of p(x) = ± 1, ± 7, ± 1/2 and ± 7/2

Putting x = 1 in p(x), We get

➝ p(1) = 2(1)³ - 13(1)² + 20(1) - 7

➝ p(1) = 2 - 13 + 20 - 7

➝ p(1) = 2 - 13 + 13

✤ Solution:

Let p(x) = 2x³ - 13x² + 20x - 7 be the given polynomial. The constant term is -7 are ± 1, ± 7. But, the factor of coefficient of x³ is 2.

So,

Possible rational roots of p(x) = ± 1, ± 7, ± 1/2 and ± 7/2

Putting x = 1 in p(x), We get

➝ p(1) = 2(1)³ - 13(1)² + 20(1) - 7

➝ p(1) = 2 - 13 + 20 - 7

➝ p(1) = 2 - 13 + 13

➝ p(1) = 2 not equals to 0

So, Let's try putting x = 1/2 in p(x),

➝ 2(1/2)³ - 13(1/2)² + 20(1/2) - 7

➝ 2 × 1/8 - 13/4 + 10 ‐ 7

➝ 1/4 - 13/4 + 3

➝ -12/4 + 3

➝ 0

|| (x - 1/2) is a factor of p(x) by factor's theoram ||

➝ (x - 1/2) = (2x - 1/2)

➝ So, (2x - 1) is also the factor.

Now arranging to get (2x - 1) in p(x),

➝ 2x³ - 13x² + 20x - 7

➝ 2x³ - x² - 12x² + 6x + 14x - 7

➝ x²(2x - 1) - 6x(2x - 1) + 7(2x - 1)

➝ (2x - 1)(x² - 6x + 7)

Further factorising (x² - 6x + 7),

➝ x² - 6x + 7

By sreedhar acharya's formula,

➝ \large{ \sf{ \dfrac{ - b \pm \sqrt{b {}^{2} - 4ac } }{2a} }}

2a

−b±

b

2

−4ac

➝ \large{ \sf{ \dfrac{ 6 \pm \sqrt{ {6}^{2} - 4 \times 1 \times 7 } }{2 \times 1} }}

2×1

6±

6

2

−4×1×7

➝ \large{ \sf{ \dfrac{6 \pm \sqrt{36 - 28} }{2} }}

2

6±

36−28

➝ \large{ \sf{ \dfrac{ - 6 \pm2 \sqrt{2} }{2} }}

2

−6±2

2

➝ \large{ \sf{ 3 + \sqrt{2}}} \: \: \& \: \: 3 - \sqrt{2}3+

2

&3−

2

So,

x = 3 + √2

x = 3 - √2

Then,

➝ 2x³ - 13x² + 20x - 7

➝ (2x - 1)(x² - 6x + 7)

➝ (2x - 1)[(x - (3 + √2)][x - (3 - √2)]

➝ (2x - 1)(x - 3 - √2)(x - 3 + √2)

☀️ Hence, factorised !!

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