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3 years ago

What is the product?(x − 3)(2x2 − 5x + 1)

Mathematics

2 answers:

svetlana [45]3 years ago

8 0

Answer:

The product of $(x-3)(2x^{2}-5x+1)$ is $2x^{3}-11x^{2}+16x-3$

Step-by-step explanation:

$(x-3)(2x^{2} -5x+1)=x(2x^{2} -5x+1) - 3(2x^{2} -5x+1)$

$=>(x-3)(2x^{2}-5x+1)=2x^{3}-5x^{2}+x-6x^{2}+15x-3$

$=>(x-3)(2x^{2}-5x+1)=2x^{3}-11x^{2}+16x-3$

Thus the product of $(x-3)(2x^{2}-5x+1)$ is $2x^{3}-11x^{2}+16x-3$

slavikrds [6]3 years ago

3 0

2x^3-11x^2+16x-3

the answew is above^^^^^^^^^^^^^

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The possible values of $b$ are -2.944 and -9.055, respectively.

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From statement we know that $AB = 2\cdot BC$ . By Analytical Geometry, we use the equation of a line segment, which is an application of the Pythagorean Theorem:

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$\sqrt{(x_{B}-x_{A})^{2}+(y_{B}-y_{A})^{2}} = 2\cdot \sqrt{(x_{C}-x_{B})^{2}+(y_{C}-y_{B})^{2}}$ (1)

Where:

$x_{A}$ , $x_{B}$ , $x_{C}$ - x-Coordinates of points A, B and C.

$y_{A}, y_{B}, y_{C}$ - y-Coordinates of points A, B and C.

$(x_{B}-x_{A})^{2}+(y_{B}-y_{A})^{2} = 4\cdot (x_{C}-x_{B})^{2}+4\cdot (y_{C}-y_{B})^{2}$

Then, we expand and simplify the expression above:

$x_{B}^{2}-2\cdot x_{A}\cdot x_{B} +x_{A}^{2} +y_{B}^{2}-2\cdot y_{A}\cdot y_{B} + y_{A}^{2} = 4\cdot (x_{C}^{2}-2\cdot x_{C}\cdot x_{B}+x_{B}^{2})+4\cdot (y_{C}^{2}-2\cdot y_{C}\cdot y_{B}+y_{B}^{2})$

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If we know that $x_{A} = 5$ , $y_{A} = 6$ , $x_{B} = 1$ , $y_{B} = b$ , $x_{C} = 1$ and $y_{C} = -3$ , then we have the following expression:

$1 -10 +25 +b^{2} -12\cdot b+36 = 100 -8 +4 +36+24\cdot b +4\cdot b^{2}$

$b^{2}-12\cdot b +52 = 4\cdot b^{2}+24\cdot b +132$

$3\cdot b^{2}+36\cdot b +80 = 0$

This is a second order polynomial, which means the existence of two possible real solutions. By Quadratic Formula, we have the following y-coordinates for point B:

$b_{1} \approx -2.944$ , $b_{2} \approx -9.055$