Question

please answer these questions :-​

Answer 1

Answer:

Q2

1) (x + 7)

2) (x + 4)

Q3

1) x = 1/2

2) x = -5/3

Step-by-step explanation:

Q2

1) $(x^2 - x - 42) / (x + 6)$

First, factor the trinomial:

$x^2 - x - 42 = (x + 6)(x - 7)$

Then, divide by the binomial, which completely cancels with the first factored term of the trinomial, (x + 6):

$(x + 6)(x - 7) / (x + 6)$

and you are left with:

x + 7 as the answer.

2) $(x^2 + 7x + 12) / (x + 3)$

Factor again:

$x^2 + 7x + 12 = (x + 3)(x + 4)$

and cancel:

$(x + 3)(x + 4) / (x + 3)$

and you are left with:

x + 4 as the answer.

Q3

1) $(x - 2)/1 + 5x = \frac{1}{2}$

First, get rid of the division by 1:

$x - 2 + 5x = \frac{1}{2}$

Combine like terms:

$6x - 2 = \frac{1}{2}$

then move the -2 to the other side:

$6x = 2 + \frac{1}{2}$

and divide by 6:

$x = \frac{3}{2} / 6$

$x = \frac{6}{4} / 6$

so you get:

$x = \frac{1}{2}$ as the answer.

2) $[(x^2 - (x-2)(x-2)] / (x + 3) = \frac{1}{2}$

Treat it like a fraction:

$\frac{(x^2 - (x-2)(x-2)} {x + 3} = \frac{1}{2}$

Cross multiply:

$2(x^2 - (x-2)^2) = 1(x + 3)$

Multiply out and solve to get x = -5/3.

Answer 2

Q. 2 ,Using factor method, divide the following polynomial by a binomial

i) x²-x-42 by x+ 6

Given:

x2 - x - 42

Divisor = x + 6

Calculation:

x2 - x - 42 ----(1)

By using remainder theorem,

⇒ x + 6 = 0

⇒ x = -6

From equation (1)

⇒ x2 - x - 42 = (-6)2 - (-6) - 42

⇒ 36 + 6 - 42

⇒ 0

∴ The remainder is 0.

ii) x²+ 7x + 12 by x +3

Let us first factorize the numerator of the given division

$\frac{x {}^{2} + 7x + 12}{x + 3}$

as follows:

x²+7x+12

=x² +4x+3x+12

=x(x+4)+3(x+4)

=(x+3)(x+4)

Since we have to divide

$\frac{x {}^{2} + 7x + 12}{x + 3}$

, therefore, we will divide

$\frac{(x + 3)(x + 4) }{x + 3}$

as shown below:

$\frac{(x + 3)(x + 4)}{x + 3} = x + 4$

Hence, $\frac{x {}^{2} + 7x + 12}{x + 3} = x + 4$