Find the constant $a$ such that\[(x^2 - 3x + 4)(2x^2 +ax + 7) = 2x^4 -11x^3 +30x^2 -41x +28.\]

Mathematics

1 answer:

Colt1911 [192]3 years ago

4 0

The value of constant a is -5

Further explanation:

We will use the comparison of co-efficient method for finding the value of a

So,

Given

$(x^2 - 3x + 4)(2x^2 +ax + 7)\\= x^2((2x^2 +ax + 7)-3x((2x^2 +ax + 7)+4(2x^2 +ax + 7)\\= 2x^4+ax^3+7x^2-6x^3-3ax^2-21x+8x^2+4ax+28\\Combining\ alike\ terms\\=2x^4+ax^3-6x^3+7x^2-3ax^2+8x^2-21x+4ax+28\\= 2x^4 +(a-6)x^3+(15-3a)x^2-(21-4a)x+28\\$

As it is given that

(x^2 - 3x + 4)(2x^2 +ax + 7) = 2x^4 -11x^3 +30x^2 -41x +28

In this case, co-efficient of variables will be equal, so we can compare the coefficients of x^3, x^2 or x

Comparing coefficient of x^3

$a-6 = -11\\a = -11+6\\a = -5$

So the value of constant a is -5

Keywords: Polynomials, factorization

Learn more about factorization at:

#LearnwithBrainly

You might be interested in

A major cab company in Chicago has computed its mean fare from O'Hare Airport to the Drake Hotel to be $28.75 , with a standard

bulgar [2K]

Answer:

Following are the solution to the given choices:

Step-by-step explanation:

Using chebyshev's theorem :

$\to P(|(x-\mu)|\leq k \sigma)\geq 1- \frac{1}{k^2}\\\\here \\ \to \mu=28.75\\\\ \to \sigma=4.44$

In point a)

$\to |(x-\mu)|= |20.17-28.75|=8.58 and \sigma=4.44 \\\\so, \\k= \frac{ |(x-\mu)| }{\sigma}$

$= \frac{8.58}{4.44} \\\\ =1.9 \\\\ =2$

$value= (1- \frac{1}{k^2}) \times 100 \% =75\%$

In point b)

$\to (1- \frac{1}{k^2}) \times 100 \% =84\% \\\\\to (1- \frac{1}{k^2})=0.84 \\\\\to \frac{1}{k^2} =0.16 \\\\\to \frac{1}{k}=0.4\\\\ \to k=2.5 \\\\\to |(x-\mu)| \leq k \sigma \\\\= |(x-\mu)|\leq 2.5 \times 4.44 \\\\ = |(x-\mu)|\leq 1.11 \\\\ = (28.75\pm 11.1) \\\\\to \text{fares lies between}(17.65,39.85)$