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Brilliant_brown [7]
3 years ago
9

Simplify the expression to a polynomial in standard form: (4x-3)(-2x^2-7x-5)

Mathematics
1 answer:
Brums [2.3K]3 years ago
6 0

How to solve your problem

(4−3)(−22−7−5)

(4x−3)(−2x2−7x−5)(4x-3)(-2x^{2}-7x-5)(4x−3)(−2x2−7x−5)

Simplify

1

Distribute

(4−3)(−22−7−5)

(4x−3)(−2x2−7x−5){\color{#c92786}{(4x-3)(-2x^{2}-7x-5)}}(4x−3)(−2x2−7x−5)

4(−22−7−5)−3(−22−7−5)

2

distribute

4(−22−7−5)−3(−22−7−5)

4x(−2x2−7x−5)−3(−2x2−7x−5){\color{#c92786}{4x(-2x^{2}-7x-5)}}-3(-2x^{2}-7x-5)4x(−2x2−7x−5)−3(−2x2−7x−5)

−83−282−20−3(−22−7−5)

3

−83−282−20−3(−22−7−5)

−8x3−28x2−20x−3(−2x2−7x−5)-8x^{3}-28x^{2}-20x{\color{#c92786}{-3(-2x^{2}-7x-5)}}−8x3−28x2−20x−3(−2x2−7x−5)

−83−282−20+62+21+15

4

Solution

−83−222++15

I know this looks like a lot but its just how you solve your problem.

therefor, your answer is

Solution

−83−222++15

<em>Hope this helps!</em>

<em>Have a great day!</em>

<em>-Hailey</em>

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Answer:

The quadratic polynomial with integer coefficients is y = 81\cdot x^{2}-36\cdot x -19.

Step-by-step explanation:

Statement is incorrectly written. Correct form is described below:

<em>Find a quadratic polynomial with integer coefficients which has the following real zeros: </em>x = \frac{2}{9}\pm \frac{\sqrt{23}}{9}<em>. </em>

Let be r_{1} = \frac{2}{9}+\frac{\sqrt{23}}{9} and r_{2} = \frac{2}{9}-\frac{\sqrt{23}}{9} roots of the quadratic function. By Algebra we know that:

y = (x-r_{1})\cdot (x-r_{2}) = x^{2}-(r_{1}+r_{2})\cdot x +r_{1}\cdot r_{2} (1)

Then, the quadratic polynomial is:

y = x^{2}-\frac{4}{9}\cdot x -\frac{19}{81}

y = 81\cdot x^{2}-36\cdot x -19

The quadratic polynomial with integer coefficients is y = 81\cdot x^{2}-36\cdot x -19.

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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