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3 years ago

Given the trinomial, what is the value of the coefficient B in the factored form?

Mathematics

2 answers:

Ivenika [448]3 years ago

8 0

Answer:

B = 6

Step-by-step explanation:

2x^2 + 4xy − 48y^2

Factor out 2

2(x^2 + 2xy − 24y^2)

What 2 numbers multiply to -24 and add to 2

-4 *6 = -24

-4+6 = 2

2 ( x+6y)( x-4y)

dalvyx [7]3 years ago

6 0

Answer:

$\huge\boxed{B=6}$

Step-by-step explanation:

They are two way to solution.

<h2>METHOD 1:</h2>

Factor the polynomial on the left side of the equation:

$2x^2+4xy-48y^2=2(x^2+2xy-24y^2)=2(x^2+6xy-4xy-24y^2)\\\\=2\bigg(x(x+6y)-4y(x+6y)\bigg)=2(x+6y)(x-4y)$

Therefore:

$2x^2+4xy-48y^2=2(x+By)(x-4y)\\\Downarrow\\2(x+6y)(x-4y)=2(x+By)(x-4y)\to\boxed{\bold{B=6}}$

<h2>METHOD 2:</h2>

Multiply everything on the right side of the equation using the distributive property and FOIL:

$2(x+By)(x-4y)=\bigg((2)(x)+(2)(By)\bigg)(x-4y)\\\\=(2x+2By)(x-4y)=(2x)(x)+(2x)(-4y)+(2By)(x)+(2By)(-4y)\\\\=2x^2-8xy+2Bxy-8By^2=2x^2+(2B-8)xy-8By^2$

Compare polynomials:

$2x^2+4xy-48y^2=2x^2+(2B-8)xy-8By^2$

From here we have two equations:

$2B-8=4\ \text{and}\ -8B=-48$

$1)\\2B-8=4$ <em>add 8 to both sides</em>

$2B=12$ <em>divide both sides by 2</em>

$B=6$

$2)\\-8B=-48$ <em>divide both sides by (-8)</em>

$B=6$

The results are the same. Therefore B = 6.

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r-ruslan [8.4K]

The answer would be four

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3 years ago

Read 2 more answers

Create a list of steps, in order, that will solve the following equation 1/4(x+5)^2-1=3

julia-pushkina [17]

Answer:

either x = -1, or x = - 9

Step-by-step explanation:

given $\frac{1}{4} (x+5)^{2} -1=3$

add one to both sides $\frac{1}{4} (x+5)^{2} =4$

im not sure which one comes first. either you multiply both the x and the 5 in the parenthesis by $\frac{1}{4}$ . or you square x and 5

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation : 1/4*(x+5)^2-(4)=0

Step by step solution :

Step 1 :

1 /4

Equation at the end of step 1 :

(— • (x + 5)2) - 4 = 0

Step 2 :

Equation at the end of step 2 :

(x + 5)2

———————— - 4 = 0

Step 3 :

Rewriting the whole as an Equivalent Fraction :

3.1 Subtracting a whole from a fraction

Rewrite the whole as a fraction using 4 as the denominator :

4 4 • 4

4 = — = ———

1 4

Equivalent fraction : The fraction thus generated looks different but has the same value as the whole

Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator

Adding fractions that have a common denominator :

3.2 Adding up the two equivalent fractions

Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

(x+5)2 - (4 • 4) x2 + 10x + 9

——————— ——————— = ——

4 4

Trying to factor by splitting the middle term

3.3 Factoring x2 + 10x + 9

The first term is, x2 its coefficient is 1 .

The middle term is, +10x its coefficient is 10 .

The last term, "the constant", is +9

Step-1 : Multiply the coefficient of the first term by the constant 1 • 9 = 9

Step-2 : Find two factors of 9 whose sum equals the coefficient of the middle term, which is 10 .

-9 + -1 = -10

-3 + -3 = -6

-1 + -9 = -10

1 + 9 = 10 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, 1 and 9

x2 + 1x + 9x + 9

Step-4 : Add up the first 2 terms, pulling out like factors :

x • (x+1)

Add up the last 2 terms, pulling out common factors :

9 • (x+1)

Step-5 : Add up the four terms of step 4 :

(x+9) • (x+1)

Which is the desired factorization

Equation at the end of step 3 :

(x + 9) • (x + 1)

————————————————— = 0

Step 4 :

When a fraction equals zero :

4.1 When a fraction equals zero ...

Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.

Now,to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator.

Here's how:

(x+9)•(x+1)

——————————— • 4 = 0 • 4

Now, on the left hand side, the 4 cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :

(x+9) • (x+1) = 0

Theory - Roots of a product :

4.2 A product of several terms equals zero.

When a product of two or more terms equals zero, then at least one of the terms must be zero.

We shall now solve each term = 0 separately

In other words, we are going to solve as many equations as there are terms in the product

Any solution of term = 0 solves product = 0 as well.

Solving a Single Variable Equation :

4.3 Solve : x+9 = 0

Subtract 9 from both sides of the equation :

x = -9

Solving a Single Variable Equation :

4.4 Solve : x+1 = 0

Subtract 1 from both sides of the equation :

x = -1

Supplement : Solving Quadratic Equation Directly

Solving x2+10x+9 = 0 directly

Earlier we factored this polynomial by splitting the middle term. let us now solve the equation by Completing The Square and by using the Quadratic Formula

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3 years ago

1. Prove or give a counterexample for the following statements: a) If ff: AA → BB is an injective function and bb ∈ BB, then |ff

Fantom [35]

Answer:

a) False. A = {1}, B = {1,2} f: A ⇒ B, f(1) = 1

b) True

c) True

d) B = {1}, A = N, f: N ⇒ {1}, f(x) = 1

Step-by-step explanation:

a) lets use A = {1}, B = {1,2} f: A ⇒ B, f(1) = 1. Here f is injective but 2 is an element of b and |f−¹({b})| = 0., not 1. This statement is False.

b) This is True. If A were finite, then it can only be bijective with another finite set with equal cardinal, therefore, B should be finite (and with equal cardinal). If A were not finite but countable, then there should exist a bijection g: N ⇒ A, where N is the set of natural numbers. Note that f o g : N ⇒ B is a bijection because it is composition of bijections. This, B should be countable. This statement is True.

c) This is true, if f were surjective, then for every element of B there should exist an element a in A such that f(a) = b. This means that f−¹({b}) has positive cardinal for each element b from B. since f⁻¹(b) ∩ f⁻¹(b') = ∅ for different elements b and b' (because an element of A cant return two different values with f). Therefore, each element of B can be assigned to a subset of A (f⁻¹(b)), with cardinal at least 1, this means that |B| ≤ |A|, and as a consequence, B is finite.

b) This is false, B = {1} is finite, A = N is infinite, however if f: N ⇒ {1}, f(x) = 1 for any natural number x, then f is surjective despite A not being finite.

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3 years ago

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Galina-37 [17]

Answer:

Step-by-step explanation:

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