Question

find all the zeroes of
${ 6x}^{2} - 27x { + 13x}^{2} \: + 63 {x \: - 6}3$
if we have two of its zeros as
$\sqrt{7 \div 3 \: and - \sqrt{7 \div 3} }$
​

Answer 1

Answer:

${2269 + - 3 { \frac{1555552222}{ { {?}^{ \\ } }^{?} } }^{?} }^{?}$

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Answer 2

Answer:

➡ANSWER

45⇒±1,±3,±5,±9,±15,±45

If we put x = 1 in p(x)

p(1)=2(1)

4

−7(1)

3

−13(1)

2

+63(1)−45

p(1)=2−7−13+63−45=65−65=0

∴x=1 or x - 1 is a factor of p(x).

Similarly if we put x = 3 in p(x)

p(3)=2(3)

4

−7(3)

3

−13(3)

2

+63(3)−45

p(3)=162−189−117+189−45=162−162=0

Hence, x = 3 or (x - 3) = 0 is the factor of p(x).

p(x)=2x

4

−7x

3

−13x

2

+63x−45

∴p(x)=2x

3

(x−1)−5x

2

(x−1)−18x(x−1)+45(x−1)

⇒p(x)=(x−1)(2x

3

−5x

2

−18x+45)

⇒p(x)=(x−1)(2x

3

−5x

2

−18x+45)

⇒p(x)=(x−1)[2x

2

(x−3)+x(x−3)−15(x−3)]

⇒p(x)=(x−1)(x−3)(2x

2

+x−15)

⇒p(x)=(x−1)(x−3)(2x

2

+6x−5x−15)

⇒p(x)=(x−1)(x−3)[2x(x+3)−5(x+3)]

⇒p(x)=(x−1)(x−3)(x+3)(2x−5)......

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