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qaws [65]
2 years ago
8

An x-method chart shows the product a c at the top of x and b at the bottom of x. Above the chart is the expression a x squared

b x c. Factor the trinomial 7x2 – 3x – 4. Which pair of numbers has a product of ac and a sum of b? What is the factored form of the trinomial?.
Mathematics
2 answers:
MaRussiya [10]2 years ago
8 0

Factored form of an expression is writing it in the terms of multiplication of factors. The factors of given expression are: (7x+4)  and (x-1)

<h3>How to find the factors of a quadratic expression?</h3>

If the given quadratic expression is of the form ax^2 + bx + c,

then its factored form is obtained by two numbers alpha( α ) and beta( β) such that:

b = \alpha + \beta \\ ac =\alpha - \beta

The given expression is 7x^2 - 3x - 4

Comparing it with the standard form of quadratic expression ax^2 + bx + c, we get: a = 7, b = -3, c = -4

ac = -28

-28 = -2 times 14

-28 = -7 times 4 (-7 + 4 = -3 = b)

Thus, \alpha = -7, \beta = 4

Thus, we get:

7x^2 - 3x - 4 = 7x^2 - 7x + 4x - 4 = 7x(x-1) + 4(x-1)  = (7x+4)(x-1)\\7x^2  - 3x  - 4 = (7x+4)(x-1)

Thus, The factors of given expression are: (7x+4)  and (x-1)

Learn more about factor of quadratic equation here:

brainly.com/question/26675692

tensa zangetsu [6.8K]2 years ago
7 0

Answer:

(x+2) (x+8)

2 nn 8

Step-by-step explanation:

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timama [110]

Use the change-of-basis identity,

\log_x(y) = \dfrac{\ln(y)}{\ln(x)}

to write

xyz = \log_a(bc) \log_b(ac) \log_c(ab) = \dfrac{\ln(bc) \ln(ac) \ln(ab)}{\ln(a) \ln(b) \ln(c)}

Use the product-to-sum identity,

\log_x(yz) = \log_x(y) + \log_x(z)

to write

xyz = \dfrac{(\ln(b) + \ln(c)) (\ln(a) + \ln(c)) (\ln(a) + \ln(b))}{\ln(a) \ln(b) \ln(c)}

Redistribute the factors on the left side as

xyz = \dfrac{\ln(b) + \ln(c)}{\ln(b)} \times \dfrac{\ln(a) + \ln(c)}{\ln(c)} \times \dfrac{\ln(a) + \ln(b)}{\ln(a)}

and simplify to

xyz = \left(1 + \dfrac{\ln(c)}{\ln(b)}\right) \left(1 + \dfrac{\ln(a)}{\ln(c)}\right) \left(1 + \dfrac{\ln(b)}{\ln(a)}\right)

Now expand the right side:

xyz = 1 + \dfrac{\ln(c)}{\ln(b)} + \dfrac{\ln(a)}{\ln(c)} + \dfrac{\ln(b)}{\ln(a)} \\\\ ~~~~~~~~~~~~+ \dfrac{\ln(c)\ln(a)}{\ln(b)\ln(c)} + \dfrac{\ln(c)\ln(b)}{\ln(b)\ln(a)} + \dfrac{\ln(a)\ln(b)}{\ln(c)\ln(a)} \\\\ ~~~~~~~~~~~~ + \dfrac{\ln(c)\ln(a)\ln(b)}{\ln(b)\ln(c)\ln(a)}

Simplify and rewrite using the logarithm properties mentioned earlier.

xyz = 1 + \dfrac{\ln(c)}{\ln(b)} + \dfrac{\ln(a)}{\ln(c)} + \dfrac{\ln(b)}{\ln(a)} + \dfrac{\ln(a)}{\ln(b)} + \dfrac{\ln(c)}{\ln(a)} + \dfrac{\ln(b)}{\ln(c)} + 1

xyz = 2 + \dfrac{\ln(c)+\ln(a)}{\ln(b)} + \dfrac{\ln(a)+\ln(b)}{\ln(c)} + \dfrac{\ln(b)+\ln(c)}{\ln(a)}

xyz = 2 + \dfrac{\ln(ac)}{\ln(b)} + \dfrac{\ln(ab)}{\ln(c)} + \dfrac{\ln(bc)}{\ln(a)}

xyz = 2 + \log_b(ac) + \log_c(ab) + \log_a(bc)

\implies \boxed{xyz = x + y + z + 2}

(C)

6 0
2 years ago
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