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Olegator [25]
3 years ago
9

Factor the polynomial 4x^4-20x^2-3x^2+15by grouping. Which is the resulting expression

Mathematics
2 answers:
Sindrei [870]3 years ago
5 0

Your answer appears to be (x^2-5)(4x^2-3) and heres why:

Steps\\$4x^4-20x^2-3x^2+15$\\\\Step 1. Factor 4x^4-20\\\\4x^4-20x^2\\\mathrm{Apply\:exponent\:rule}:\quad \:a^{b+c}=a^ba^c\\x^4=x^2x^2\\=4x^2x^2-20x^2\\\\\mathrm{Rewrite\:as}\\=4x^2x^2-5\cdot \:4x^2\\\\\mathrm{Factor\:out\:common\:term\:}4x^2\\=4x^2\left(x^2-5\right)\\\\$Step 2. \mathrm{Factor}\:-3x^2+15:\quad 3\left(-x^2+5\right)$\\-3x^2+15\\\\\mathrm{Rewrite\:as}\\=-3x^2+3\cdot \:5\\\\\mathrm{Factor\:out\:common\:term\:}3\\=3\left(-x^2+5\right)\\\\

Step\ 3.\\ $=4x^2\left(x^2-5\right)+3\left(-x^2+5\right)$\\\\\mathrm{Rewrite\:as}\\=4\left(x^2-5\right)x^2-3\left(x^2-5\right)\\\\\mathrm{Factor\:out\:common\:term\:}\left(x^2-5\right)\\=\left(x^2-5\right)\left(4x^2-3\right)

<h3>Now, if you have anymore questions and would like to see more answers like this, please make sure to Like, Comment, and Rate! Also, make sure to send me a friend request. I'm always up for a challenge!</h3><h2>Thanks!</h2>

ps. could you please mark me as Brainlyest? I kinda need it to help maintain and to reach the next rank... thx

Arisa [49]3 years ago
5 0

Answer:

\boxed{\bold{\left(x^2-5\right)\left(4x^2-3\right)}}

Step By Step Explanation:

Factor \bold{4x^4-20x^2}

\bold{4x^2\left(x^2-5\right)}

Factor \bold{-3x^2+15}

\bold{3\left(-x^2+5\right)}

Rewrite Equation

\bold{4x^2\left(x^2-5\right)+3\left(-x^2+5\right)}

Rewrite As...

\bold{4\left(x^2-5\right)x^2-3\left(x^2-5\right)}

Factor Out Common Term \bold{\left(x^2-5\right)}

\bold{\left(x^2-5\right)\left(4x^2-3\right)}

\boxed{\bold{\black{Mordancy}}}

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Delta makes 12-volt car batteries. These batteries are known to be normally
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Answer:

The probability that Delta car batteries last between three and four years

P(36≤X≤48) = 0.5188

The percentage of that Delta car batteries last between three and four years

P(3≤X≤4) = 52%

Step-by-step explanation:

<u><em>Step(i):-</em></u>

<em>Given that the sample size n =12 -volt car batteries</em>

<em>Let  'X' be a Random variable in a normal distribution</em>

<em>Given that mean of the normal distribution = 45 months</em>

<em>Given that the Standard deviation of the normal distribution = 8months</em>

<u><em>Step(ii):-</em></u>

Let  X₁ = 3 years = 12 × 3 = 36 months

Z_{1} = \frac{x_{1} -mean}{S.D} = \frac{36-45}{8} = -1.125

Let X₂ = 4 years  = 12 × 4 = 48 months

Z_{2} = \frac{x_{2} -mean}{S.D} = \frac{48-45}{8} = 0.375

<u><em>Step(iii)</em></u>:-

The probability that Delta car batteries last between three and four years

P(36≤X≤48) = P(-1.125≤Z≤0.375)

                   = P(Z≤0.375) - P(Z≤-1.125)

                   = 0.5 +A(0.375) - (0.5-A(1.125)

                   = 0.5 + 0.1480 - (0.5 -0.3708)

                  = 0.1480 + 0.3708

                 = 0.5188

<u><em>Final answer:-</em></u>

The probability that Delta car batteries last between three and four years

P(36≤X≤48) = 0.5188

The percentage of that Delta car batteries last between three and four years

P(3≤X≤4) = 52%

<em />

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