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2 years ago

Expand and simplfy (x+3)(x-3)(3x+2)

Mathematics

1 answer:

tester [92]2 years ago

4 0

Answer:

${3x}^{3} + {2x}^{2} - 27x - 18$

Step-by-step explanation:

1) Use difference of squares: a² - b² = (a + b) (a - b).

$( {x}^{2} - {3}^{2} )(3x + 2)$

2) Simplify 3² to 9.

$( {x}^{2} - 9)(3x + 2)$

3) Use The FOIL method: (a + b) (c + d) = ac + ad + bc + bd.

${3x}^{3} + {2x}^{2} - 27x - 18$

Therefor, the answer is 3x³ + 2x² + -26x -18.

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2 years ago

write a polynomial function of least degree with integral coefficients that has the given zeros. -(1/3), -i

inessss [21]

Answer:

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

Step-by-step explanation:

If a real number $-\frac{1}{3}$ is a zero of polynomial function, then

$x-\left(-\dfrac{1}{3}\right)=x+\dfrac{1}{3}$

is the factor of this function.

If a complex number $-i$ is a xero of the polynomial function, then the complex number $i$ is also a zero of this function and

$x-(-i)=x+i\ \text{ and }\ x-i$

are two factors of this function.

So, the function of least degree is

$f(x)=\left(x+\dfrac{1}{3}\right)(x+i)(x-i)=\left(x+\dfrac{1}{3}\right)(x^2-i^2)=\\ \\ =\left(x+\dfrac{1}{3}\right)(x^2+1)=\dfrac{1}{3}(3x+1)(x^2+1)=\dfrac{1}{3}(3x^3+x^2+3x+1)$

If the polynomial function must be with integer coefficients, then it has a form

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

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3 years ago