How do I factor 2x^2+7x+3?
always try to split the middle term such that product of first and last terms is equal to product of split termswe can split 7 as 6+1, 5+2, 4+3 and first and last term multiplies into 6(3x2)we can split 7x as 6x+xhence 2x^2+7x+1= 2x^2+6x+x+3=2x(x+3)+1(x+3)=(2x+1)(x+3)
How to factor 2x^2-3x-27?
its factorable (2x-9)(x+3) you can check by using the discriminant B^2-4AC if the answer is prefect square thain you car factor it
How do you factor 2x^3-x^2-145x-72?
1. factor an X X(2X^2-X-145-72) 2. combine like terms X(2X^2-X-73) 3. now all you have to do is factor that equation. It should look like the following (2X-?)(X+? )
How can I factor 2x^2-5x-7?
2x^2–5x-7 = 0The product of 2 and 7 is 14. The difference of 7 and 2 is 5.So we write the equation as2x^2+2x-7x-7 = 02x(x+1)-7(x+1) = 0(2x-7)(x+1) = 0 are the factors desired.
How to factor 2x^3-6x+4?
(2x -2)(x - 2)
Factor 2x^3 - 5x^2 - 3x = 0 ... PLEASE!!!?
x( 2x2 - 5x -3) = 0 (factoring out an x) x (2x + 1) (x -3) = 0 ( factoring the quadratic) Set each one equal to 0 So x = 0.....or 2x + 1 = 0 ......or x - 3 = 0 ....................2x = -1..............x = 3 ......................x = -1/2 so x = -1/2, 0, 3
How do I factor [math]2x^3+3x^2+4[/math]?
For high degree polynomials, a good starting point is to check possible rational roots. The factors of 4 are 1, 2, and 4. The factors of 2 are 1 and 2. Therefore: the possible rational roots would be the possible factors of 4 divided by the possible factors of 2:[math]\pm \frac{4}{2}, \pm \frac{2}{2}, \pm \frac{1}{2}, \pm \frac{4}{1}, \pm \frac{2}{1}, \pm \frac{1}{1}[/math]Simplifying and eliminating duplicates, that gives us:[math]\pm 4, \pm 2, \pm 1, \pm \frac{1}{2}[/math]Therefore, you plug in those 8 values to see if any of them are a 0 for the polynomial. After doing so, you can find that -2 is a zero. That means one of the factors of the polynomial is [math](x+2)[/math]Then you perform polynomial division to find out what remains after factoring that out. You will get an answer that looks like [math](x+2)(ax^2+bx+c)[/math] I will let you figure out what the values of [math]a, b,[/math] and [math]c[/math] are.
How would you factor 2x^2 + 5x + 3?
1. You need to find two numbers whose product is 2 * 3 (the first and last coefficients) and whose sum is 5 (the second coefficient) The numbers are 2 and 3. 2. Split the 5x into two parts using the numbers you found. 2x^2 + 5x + 3 = 2x^2 + 2x + 3x + 3 3. Now, by grouping two terms at a time, factor out the largest factor: 2x^2 + 2x + 3x + 3 = 2x(x + 1) + 3(x + 1) 4. Then, factor out the common factor (x+1) (x+1)(2x +3) If you do not understand step 4, think of it this way: let y = (x=1) 2xy + 3y = y(2x+3)
What is the value of c if x+2 is a factor of 2x^3+2x^2-2cx+4?
2x³+2x²-2cx+4First, we can divide everything by 2:2x³+2x²-2cx+4=2(x³+x²-cx+2)If x+2 is a factor, then we divide by x+2:2(x³+x²-cx+2)=2(x³+x²-cx+2)/(x+2)=2(x+2)(x²-x+1)If c=1, then (x+2) would be a factor……
What is the value of a and b if x^2-4 is a factor of ax^4 +2x^3 - 3x^2 +bx?
GIVEN: p(x) = ax^4+2x^3–3x²+bxg(x) = x²-4Where p(x) & g(x) are polynomials in variable ‘x'& here g(x) is a factor of p(x)That means p(x) is exactly divisible by g(x)Or p(x) is exactly divisible by factors of g(x)Now, factors of g(x) = x²-4 = (x+2)(x-2)So, p(x) is exactly divisible by (x+2) & (x-2)That means, if p(x) is divided by (x+2) & then by (x-2) , the remainder has to be zero.So now we find out the remainder in each case by remainder theorem:If p(x) ÷(x+2) , the ramainder = p(-2)ie, p(x)= ax^4+2x^3–3x²+bx is divided by (x+2)Remainder= p(-2)= 16a - 16 -12 -2b =0………(1)& if p(x)= ax^4+2x^3–3x²+bx is divided by (x-2)Remainder= p(2)= 16a+16–12+2b=0……….(2)eq(1) +eq(2)=> 32a-24=0=>32a =24So a= 24/32 = 3/4………(3)Now, eq(1) _ eq(2)=> -32 -4b =0=> 4b = -32=> b= -8…………(4)ANS a= 3 /4b= —8