Math, asked by divyamanisingh860, 1 month ago

1. f(x) = x3 - 6x2 +11x - 6; g(x) = x - 3
2. f(x) = 3x* + 17x + 9x? - 7x - 10; g(x) = x + 5
3. f(x) = x + 3x - x - 3x² + 5x + 15, g(x) = x + 3
4. f(x) = x2 - 6x2 - 19x +84, (x) = x - 7
5. f(x) = 3x + x2 – 20x + 12, g(x) = 3x - 2
6. f(x) = 2x3 - 9x² + x + 12, 8(x) = 3 - 2x
f(x) = x® - 6x +11% -6,8(x) = x2 – 3x + 2​

Answers

Answered by shivakumarswasti
2

1.According to Factor theorem, if (x - a) is a polynomial factor f(x), then f(a) = 0

Let f(x) = x^{3}-6 x^{2}+11 x-6f(x)=x

3

−6x

2

+11x−6

Let us check if (x - 1) is the factor of f(x),

Then,

f(1) = 1^{3}-6\left(1^{2}\right)+11(1)-6=1-6+11-6=0f(1)=1

3

−6(1

2

)+11(1)−6=1−6+11−6=0

Therefore (x-1) is a factor of f(x)

Let us check for the other factors

Hence,

f(x)=(x-1)\left(x^{2}-5 x+6\right)f(x)=(x−1)(x

2

−5x+6)

x^{2}-5 x+6=x^{2}-2 x-3 x+6x

2

−5x+6=x

2

−2x−3x+6

=x(x-2)-3(x-2)=x(x−2)−3(x−2)

= (x - 2)(x - 3)=(x−2)(x−3)

f(x) = (x - 1)(x - 2)(x - 3)f(x)=(x−1)(x−2)(x−3)

Therefore, 1, 2, 3 are the factors of f(x)

2.According to the problem:-

Given that:

f(x)=3x

4

+17x

3

+9x

2

−7x−10

and

g(x)=x+5

Using the factor theorem, g(x) is a factor of polynomial f(x)

Therefore,

x+5=0

x=−5

Hence,

f(−5)=3(−5)

4

+17(−5)

3

+9(−5)

2

−7(5)−10

Implies that

=3×625+17×(−125)+225+35−10

Implies that

=1875−2125+250

=0

Hence the g(x) is the factor of polynomial f(x)

3.Given:-

p(x)=x

3

−3x

2

+5x−3

g(x)=x

2

−2

p(x)÷g(x)

(x

3

−3x

2

+5x−3)÷(x

2

−2)

x

2

−2

x

3

−3x

2

+5x−3

=(x−3)+

(x

2

−2)

(7x−9)

Quotient =(x−3)

Remainder =(7x−9)

4.The chosen topic is not meant for use with this type of problem. Try the examples below.

f

(

x

)

=

5

x

3

+

6

f

(

x

)

=

6

4

x

f

(

x

)

=

4

x

3

5.By using factor theorem

(3x−2)=0

3x=2

x=2/3

now, let f(x)=3x

3

+x

2

−20x+12

f(2/3)=3(2/3)

3

+(2/3)

2

−20(2/3)+12

f(2/3)=3(8/27)+(4/9)−(40/3)+12

f(2/3)=(8/9)+(4/9)−(40/3)+12

f(2/3)=(12/9)−(40/3)+12

f(2/3)=(4/3)−(40/3)+12

f(2/3)=12−(36/3)

f(2/3)=12−12

f(2/3)=0

(3x−2) is a factor of 3x

3

+x

2

−20x+12

6.y - y1 = m (x - x1)

We are already given the point (-4, 9), so we will plug those values in for x1 and y1 in the above equation.

To find the slope, m, of the tangent line, we need to find the derivative of the given function and plug in x = -4:

f'(x) = 2x + 3

f'(-4) = 2(-4) + 3 = -5, so m = -5.

Now, write the equation in point-slope form:

y - 9 = -5 (x + 4)

If you need to write the equation in slope-intercept form, solve for y:

y = -5x -11

In standard form, the equation will be: 5x + y = -11

I hope this helps. Please let me know if you have any more questions.

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