Factor completely x3 + 2x2 – 15x

Answers

Answer 1

Answer: $x^3 + 2x^2 - 15x$

first we can factor out an x
$x(x^2 + 2x - 15)$

then we find a pair of numbers that multiply to -15 and add up to 2

this would be 5 and -3
because 5 times -3 = -15
and -3 + 5 = 2

we would then get
$x(x-3)(x+5)$

this would be you answer :)

Answer 2

Answer:

Hi Lisa

x³ + 2x² - 15x

= x( x-3)(x+5)

I hope that's help:)

Have a great summer :)

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Evaluate 15 − 4 + (7 − 5)2 = __________.

Answers

26? right? cause 15 - 4 is 11, 7 - 5 is 2, 11 + 2 is 13, and 13 x 2 is 26

15-4=11 plus
(7-5)= 2
if is times 2 is 32
if is plus 2 is 15
if is minus 2 is 11
and if is divided by 2 is 6.5
(I'm saying all that because you didn't put any symbol beside the number 2)

Why can't you factor 2cosx^2+sinx-1=0 ?

Answers

$2cos^2x+sinx-1=0\n\n2(1-sin^2x)+sinx-1=0\n\n2-2sin^2x+sinx-1=0\n\n-2sin^2x+sinx+1=0\n\n-2sin^2x+2sinx-sinx+1=0\n\n-2sinx(sinx-1)-1(sinx-1)=0\n\n(sinx-1)(-2sinx-1)=0\iff sinx-1=0\ or\ -2sinx-1=0\n\nsinx=1\ or\ -2sinx=1\n\nsinx=1\ or\ sinx=-(1)/(2)\n\nx=(\pi)/(2)+2k\pi\ or\ x=-(\pi)/(6)+2k\pi\ or\ x=(7\pi)/(6)+2k\pi\ where\ k\in\mathbb{Z}$

$2cosx^2+sinx-1=2(1-sin^2x)+snx-1=\n\n=2(1-sinx)(1+sinx)-(1-sinx)=(1-sinx)[2(1+sinx)-1]=\n\n=(1-sinx)(2+2sinx-1)=(1-sinx)(1+2sinx)\n\n2cosx^2+sinx-1=0\ \ \ \Leftrightarrow\ \ \ (1-sinx)(1+2sinx)=0\n\n1-sinx=0\ \ \ \ \ or\ \ \ \ \ 1+2sinx=0\n\n1)\ \ \ 1-sinx=0\ \ \ \Rightarrow\ \ \ sinx=1\ \ \ \Rightarrow\ \ \ x= ( \pi )/(2) +2k \pi ,\ \ \ k\in I\n\n$

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Which equation is quadratic in form?6(x + 2)2 + 8x + 2 + 1 = 0
6x4 + 7x2 – 3 = 0
5x6 + x4 + 12 = 0
x9 + x3 – 10 = 0

Answers

Answer:

Option A is correct

Step-by-step explanation:

We have been given four equations and we need to tell which one of them is quadratic

Case1:

$6(x+2)^2+8(x+2)+1$

In this we will use the formula $(a+b)^2=a^2+b^2+2ab$

Here, a=x and b=2

The equation will become $6(x^2+2^2+4x)+8x+16+1$

Hence, after simplification equation will become

$6x^2+24+24x+8x+16+1$

$6x^2+32x+41$ which is a quadratic equation because quadratic equation is the equation is the equation which has degree 2.

In this equation degree is 2 hence, quadratic

Case2:

$6x^4+7x^2-3$ is not quadratic since, degree in this equation is 4 not 2

Hence, biquadratic not quadratic

Case3:

$5x^6+x^4+12$ is not a quadratic equation since, degree in this equation is 6.

Hence, not quadratic

Case4:

$x^9+x^3-10$ is not quadratic since, degree in this equation is 9

Hence, not quadratic

Therefore, Option A is correct

A quadratic equation is a polynomial with an order of two. Its general form is ax² + bx + c = 0. From the choices given, the first option seems to be the quadratic equation. Simplifying the equation gives 6x² + 18x + 27 = 0.

Which of the following statements is true?

Answers

Answer:

option C is correct.

Step-by-step explanation:

Checking equation A

$(9x^3 + 12x) + (16x^3 -4x+1)= 24x^3 +16x-1\n9x^3+16x^3+12x-4x+1=24x^3+16x-1\n25x^3+8x+1\neq 24x^3+16x-1$

Checking equation B

$(3x+2x^2-4)-(x^2+x-5)=2x+x^2-9\n3x-x+2x^2-x^2-4+5=2x+x^2-9\n2x+x^2+1\neq 2x+x^2-9$

Checking equation C

$(10x^3+2x^2-11)+(9x^2+2x-2)=10x^3+11x^2+2x-13\n10x^3+2x^2+9x^2+2x-11-2=10x^3+11x^2+2x-13\n10x^3+11x^2+2x-13=10x^3+11x^2+2x-13$

Checking equation D

$(-7x^2+12)-(5-4x^2)=11x^2-17\n-7x^2+4x^2+12-5=11x^2-17\n-3x^2+7\neq 11x^2-17$

Hence, only option C is correct.

Solve this equation: –9h – 6 + 12h + 40 = 22. A. h = 24 B. h = – 4 C. h = 4 D. h = –4/7

Answers

In the equation: -9h - 6 + 12h + 40 = 22, collecting like terms gives: -9h + 12h = 22 + 6 - 40 which gives 3h = -12. They we divide both sides of the equation by 3 to get 3h / 3 = -12 / 3 which gives h = -4. Therefore, the solution to the given equation is h = -4 (option B).

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