Question

Please help! I feel like I'm drowning :(

Answer 1

Answer:

1d = -3

2b = 2

2c = 1

3a = 3

3d = 4

Step-by-step explanation:

Polynomial 1: $x^2-8x+15$

Multiply the leading coefficient, 1, and the last term, 15. You get: 15.

Then, list out the factors of 15 and the addends of -8 until you get two of numbers that are the same:

Factors of 15: -5 * -3

Addends of -8: -5 + -3

Replace the -8x with -5x - 3x:

$x^2-5x-3x+15$

Put parentheses around the first 2 terms & last 2 terms and factor like so:

$(x^2-5x)-(3x+15)$

$x(x-5)-3(x-5)$

$(x-5)(x-3)$

Looking at the answer (ax + b)(cx + d), d would correspond with -3.

Polynomial 2: $2x^3-8x^2-24x$

First factor out the x:

$x(2x^{2}-8x-24)$

Divide the polynomial inside by 2 and place the 2 outside with the x:

$2x(x^2-4x-12)$

Then find the factors of 1*-12 and the addends of -4 and see which two numbers match:

Factors of -12: -6 * 2

Addends of -4: -6 + 2

Replace the -8x with -6x + 2x:

$2x(x^2-6x+2x-12)$

Put parentheses around the first 2 terms & last 2 terms and factor like so:

$2x((x^2-6x)+(2x-12))$

$2x(x(x-6)+2(x-6))$

$2x((x+2)(x-6))$

$2x(x+2)(x-6)$

Looking at the answer (2x)(ax + b)(cx + d), b & c would correspond with 2 & 1.

Polynomial 3: $6x^2+14x+4$

Divide the polynomial by 2:

$(2)(3x^2+7x+2)$

Find the factors of 3*2 and the addends of 7 and see which two numbers match:

Factors of 6: 6 * 1

Addends of 7: 6 + 1

Replace the 7x with 6x + x:

$(2)(3x^2+6x+x+2)$

Put parentheses around the first 2 terms & last 2 terms and factor like so:

$(2)((3x^2+6x)+(x+2))$

$(2)(3x(x+2)+(x+2))$

$(2)((3x+1)(x+2))$

$(2)(3x+1)(x+2)$

Then multiply the 2 with the (x+2) and here's your final answer:

$(3x+1)(2x+4))$

Looking at the answer (ax + b)(cx + d), a & d correspond with 3 & 4.

Hope that helps (●'◡'●)

(This took a while to write, sorry about that)