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Rasek [7]
2 years ago
15

Factor each trinomial. Then match the polynomial (term) on the left with its factored form (definition) on the right. (4 points)

Group of answer choices x2 − 4x − 12 [ Choose ] x2 + 4x − 12 [ Choose ] x2 − x − 12 [ Choose ] x2 − 7x − 12 [ Choose ] i will give 40 points
Mathematics
1 answer:
Aloiza [94]2 years ago
8 0

Answer:

Your input: factor x2+4x+3.

To factor the quadratic function x2+4x+3, we should solve the corresponding quadratic equation x2+4x+3=0.

Indeed, if x1 and x2 are the roots of the quadratic equation ax2+bx+c=0, then ax2+bx+c=a(x−x1)(x−x2).

Solve the quadratic equation x2+4x+3=0.

The roots are x1=−1, x2=−3 (use the quadratic equation calculator to see the steps).

Therefore, x2+4x+3=1(x+1)(x+3).

(x2+4x+3)=1(x+1)(x+3)

Step-by-step explanation:

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Vilka [71]

Answer:

\Large \boxed{\sf \bf \ \ \dfrac{x^2-x-6}{x^2-3x+2} \ \ }

Step-by-step explanation:

Hello, first of all, we will check if we can factorise the polynomials.

\boxed{x^2+6x+8}\\\\\text{The sum of the zeroes is -6=(-4)+(-2) and the product 8=(-4)*(-2), so}\\\\x^2+6x+8=x^2+2x+4x+8=x(x+2)+4(x+2)=(x+2)(x+4)

\boxed{x^2+3x-10}\\\\\text{The sum of the zeroes is -3=(-5)+(+2) and the product -10=(-5)*(+2), so}\\\\x^2+3x-10=x^2+5x-2x-10=x(x+5)-2(x+5)=(x+5)(x-2)

\boxed{x^2+2x-15}\\\\\text{The sum of the zeroes is -2=(-5)+(+3) and the product -15=(-5)*(+3), so}\\\\x^2+2x-15=x^2-3x+5x-15=x(x-3)+5(x-3)=(x+5)(x-3)

\boxed{x^2+3x-4}\\\\\text{The sum of the zeroes is -3=(-4)+(+1) and the product -4=(-4)*(+1), so}\\\\x^2+3x-4=x^2-x+4x-4=x(x-1)+4(x-1)=(x+4)(x-1)

Now, let's compute the product.

\dfrac{x^2+6x+8}{x^2+3x-10}\cdot \dfrac{x^2+2x-15}{x^2+3x-4}\\\\\\=\dfrac{(x+2)(x+4)}{(x+5)(x-2)}\cdot \dfrac{(x+5)(x-3)}{(x+4)(x-1)}\\\\\\\text{We can simplify}\\\\=\dfrac{(x+2)}{(x-2)}\cdot \dfrac{(x-3)}{(x-1)}\\\\\\=\large \boxed{\dfrac{x^2-x-6}{x^2-3x+2}}

So the correct answer is the first one.

Thank you.

7 0
3 years ago
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