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3 years ago

Factorization of 3x2-8x-3

Mathematics

2 answers:

choli [55]3 years ago

8 0

Your answer is (3x + 1) (x - 3)

Hope this helps.

pychu [463]3 years ago

3 0

3x² - 8x -3 = 3x² - 9x + x - 3 =

= 3x² - 9x + x - 3 = 3x* ( x - 3) + 1* (x - 3) =

= 3x * (x - 3) + 1 * (x - 3) = (3x + 1) (x - 3)

Answer (3x + 1) (x - 3)

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Find the least common multiple (LCM) of 8y^6+ 144y^5+ 640y^4 and 2y^4 + 40y^3 + 200y^2.

Mice21 [21]

Answer:

LCM = $8y^{4}(y+ 10)^{2}(y + 8)$

Step-by-step explanation:

Making factors of $8y^{6}+ 144y^{5}+ 640y^{4}$

Taking $8y^{4}$ common:

$\Rightarrow 8y^{4} (y^{2}+ 18y+ 80)$

Using factorization method:

$\Rightarrow 8y^{4} (y^{2}+ 10y + 8y + 80)\\\Rightarrow 8y^{4} (y (y+ 10) + 8(y + 10))\\\Rightarrow 8y^{4} (y+ 10)(y + 8))\\\Rightarrow \underline{2y^{2}} \times 4y^{2} \underline{(y+ 10)}(y + 8)) ..... (1)$

Now, Making factors of $2y^{4} + 40y^{3} + 200y^{2}$

Taking $2y^{2}$ common:

$\Rightarrow 2y^{2} (y^{2}+ 20y+ 100)$

Using factorization method:

$\Rightarrow 2y^{2} (y^{2}+ 10y+ 10y+ 100)\\\Rightarrow 2y^{2} (y (y+ 10) + 10(y + 10))\\\Rightarrow \underline {2y^{2} (y+ 10)}(y + 10) ............ (2)$

The underlined parts show the Highest Common Factor(HCF).

i.e. HCF is $2y^{2} (y+ 10)$ .

We know the relation between LCM, HCF of the two numbers 'p' , 'q' and the numbers themselves as:

$HCF \times LCM = p \times q$

Using equations (1) and (2): $\Rightarrow 2y^{2} (y+ 10) \times LCM = 2y^{2} \times 4y^{2}(y+ 10)(y + 8) \times 2y^{2} (y+ 10)(y + 10)\\\Rightarrow LCM = 2y^{2} \times 4y^{2}(y+ 10)(y + 8) \times (y + 10)\\\Rightarrow LCM = 8y^{4}(y+ 10)^{2}(y + 8)$

Hence, LCM = $8y^{4}(y+ 10)^{2}(y + 8)$

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Step-by-step explanation:

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Answer:

Step-by-step explanation:

Range is largest number subtracted with the smallest number. So it will be

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2 years ago

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german tiene 13 billetes en su bolsillo , algunos son de $2000 y otros de $5000 si en total German tiene $40.000 ¿Cuantos billet

Rufina [12.5K]

Answer:

hay 10 billetes de 2000

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Step-by-step explanation:

Planteamos las siguientes ecuaciones...:

x= billetes de 2000

y= billetes de 5000

ecuaciones :

1) x+y = 14

2) 2000x + 5000y =40000

despejamos primera ecuacion y remplazamos en segunda :

1) x= 14-y

2) 2000(14-y) + 5000y = 40000

28000- 2000y+5000y = 40000

3000y= 40000 - 28000

y= 4

reemplazamos este valor en primera ecuacion:

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x+ 4= 14

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ahora reemplazamo valor de X y valor de Y

2000(10) + 5000(4) =40000

40000 =40000

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hay 4 billetes de 5000

Espero que esto te ayude

8 0

3 years ago