Question

Разложение квадратный трёхчлен на линейные множители помогите пожалуйста!

Answer 1

$ax^{2}+bx+c=a(x-x_{1})(x-x_{2})\\ \\ 7x^{2}+72x+20=0\\ x^{2}+72x+140=0\\ x_{1}+x_{2}=-72\\ x_{1}x_{2}=140\\ x_{1}=-70\\ x_{2}=-2$

Т.к. мы приводили квадратное уравнение к стандартному виду разделим получившиеся корни на 7:
$x_{1}=-10\\ x_{2}=- \frac{2}{7}\\ \\ 7x^{2}+72x+20=7(x+10)(x+ \frac{2}{7} )$

Answer 2

Factoring a Quadratic Trinomial

To factor a quadratic trinomial, such as 2X^2+5X-3, into linear factors, we can use the quadratic formula or factorization by grouping.

The quadratic formula is given by: x = (-b ± √(b^2 - 4ac)) / (2a)

The factorization by grouping method involves splitting the middle term of the trinomial and then factoring by grouping.

Let's proceed with the factorization by grouping method for the given quadratic trinomial.

The given quadratic trinomial is: 2X^2+5X-3

First, we need to find two numbers that multiply to the product of the coefficient of the quadratic term and the constant term, and add to the coefficient of the linear term. In this case, the product is -6 (2 * -3) and the sum is 5.

The numbers are -1 and 6, as -1 * 6 = -6 and -1 + 6 = 5.

Now, we can rewrite the middle term using these numbers: 2X^2 - X + 6X - 3

Next, we factor by grouping: (2X^2 - X) + (6X - 3) X(2X - 1) + 3(2X - 1)

Now, we can factor out the common binomial factor: (2X - 1)(X + 3)

So, the factored form of the quadratic trinomial 2X^2+5X-3 is (2X - 1)(X + 3).

This method allows us to factor the given quadratic trinomial into linear factors.