2xsquared+x-15?

Question

Thank you for the last answer. I finally understood it. Now I need help with this one

Answer 1

There is no equal sign, which means you won't solve for "x"...you need to factor the expression.

First: multiply the 1st & 3rd term to get (- 30). find two numbers that give you (- 30) when multiplied & 1 (2nd term) when added (or subtracted). the numbers are 6 & -5

Sec: rewrite the expression with the new *middle* numbers.

2x^2 + 6x - 5x - 15

Third: group "like" terms & factor both set of parenthesis.

(2x^2 + 6x) - (5x - 15)
2x(x + 3) - 5(x + 3)

*The inner terms (terms in parenthesis) should be the same. Combine the inner terms (once) with the outer terms.

(x+3)(2x-5) or, (2x-5)(x+3)

Answer 2

Factorise 2x² + x - 15

/ For reference :
Divide the expression by 2 and it becomes x² + 1/2 X - 15/2

The product of the roots should be -15/2 (ie: a * c) and the sum of the roots should be -1/2 (i.e : -(b/a) ) /

The roots of the quadratic equation are

x = (-b ± √b² - 4ac) /2a
where a , b and c are coefficients of each of the expression
ie: a = 2 b = 1 c = -15

x = ( - (1) ± √ ( 1)² - 4(2)(-15) ) / 2(2)
x = (-1 ± √1 + 120) / 4
x =(-1 ± √121) /4
x = (-1 ± 11 )/4

x = (-1+11)/4 = 10/4 = 5/2

x = (-1-11)/4 = -12/4 = -3

/ For reference only
Sum of the roots = 5/2 + (-3) = 5/2 -3 = -1/2 and Product of the roots = 5/2 *-3 = -15/2 /

Answer 3

again the following is not a perfect square trinomial and also not a simple trinomial since the leading coeficient is not equal to 1 making it a general trinomial then can be written in the general quadratic form of :
ax^2 + Bx + C
where a is not equal to zero and a is not equal to 1....
to factor it
step 1: form the product AC so 2*-15 = -30

step 2: list the pairs of intregers with product ac
so (3*-10) , (-3 *10), (2* -15), (-2* 15), (1 * -30) , (-1, 30) , (-5 * 6), (5 * -6)

step 3: find within the factors above wich sum will result to B which is equivalent to 1 so (-5 and 6 ) satisfies step 3
because -5 + 6 = 1 = B

Step 4: replace b in the polynomial with the indicated sum of the integers obtained in step 3 and distribute X to both terms
so 2x^2 - 5x + 6x - 15

step 5: factor the four term polynomial formed in step 4, using the grouping in pairs technique
so
x(2x-5) + 3 (2x+5)

then the simplified form will be
(2x+5)*(x+3)


i hope i helped you

Answer 4

This is a quadratic equation set to equal zero ax^2 + bx +c = 0

Then use the quadratic formula to solve for x:

x = -b + or - square root (B^2 - 4AC)/2A

X will have two values.

Answer 5

2x^2 + x -15

(2x - 5) (x + 3)

2x - 5 = 0
2x = 5
x = 5/2 or 2 1/2

x + 3 = 0
x = -3

Solution set: (2 1/2, -3)

Answer 6

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

x=-5/2 ; -3