Answer:
Quadratic polynomial can be factored using the transformation ax
2
+bx+c=a(x−x
1
)(x−x
2
), where x
1
and x
2
are the solutions of the quadratic equation ax
2
+bx+c=0.
−x
2
−3x+5=0
All equations of the form ax
2
+bx+c=0 can be solved using the quadratic formula:
2a
−b±
b
2
−4ac
. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.
x=
2(−1)
−(−3)±
(−3)
2
−4(−1)×5
Square −3.
x=
2(−1)
−(−3)±
9−4(−1)×5
Multiply −4 times −1.
x=
2(−1)
−(−3)±
9+4×5
Multiply 4 times 5.
x=
2(−1)
−(−3)±
9+20
Add 9 to 20.
x=
2(−1)
−(−3)±
29
The opposite of −3 is 3.
x=
2(−1)
3±
29
Multiply 2 times −1.
x=
−2
3±
29
Now solve the equation x=
−2
3±
29
when ± is plus. Add 3 to
29
.
x=
−2
29
+3
Divide 3+
29
by −2.
x=
2
−
29
−3
Now solve the equation x=
−2
3±
29
when ± is minus. Subtract
29
from 3.
x=
−2
3−
29
Divide 3−
29
by −2.
x=
2
29
−3
Factor the original expression using ax
2
+bx+c=a(x−x
1
)(x−x
2
). Substitute
2
−3−
29
for x
1
and
2
−3+
29
for x
2
.
−x
2
−3x+5=−(x−
2
−
29
−3
)(x−
2
29
−3
)
EVALUATE
5−3x−x
2Quadratic polynomial can be factored using the transformation ax
2
+bx+c=a(x−x
1
)(x−x
2
), where x
1
and x
2
are the solutions of the quadratic equation ax
2
+bx+c=0.
−x
2
−3x+5=0
All equations of the form ax
2
+bx+c=0 can be solved using the quadratic formula:
2a
−b±
b
2
−4ac
. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.
x=
2(−1)
−(−3)±
(−3)
2
−4(−1)×5
Square −3.
x=
2(−1)
−(−3)±
9−4(−1)×5
Multiply −4 times −1.
x=
2(−1)
−(−3)±
9+4×5
Multiply 4 times 5.
x=
2(−1)
−(−3)±
9+20
Add 9 to 20.
x=
2(−1)
−(−3)±
29
The opposite of −3 is 3.
x=
2(−1)
3±
29
Multiply 2 times −1.
x=
−2
3±
29
Now solve the equation x=
−2
3±
29
when ± is plus. Add 3 to
29
.
x=
−2
29
+3
Divide 3+
29
by −2.
x=
2
−
29
−3
Now solve the equation x=
−2
3±
29
when ± is minus. Subtract
29
from 3.
x=
−2
3−
29
Divide 3−
29
by −2.
x=
2
29
−3
Factor the original expression using ax
2
+bx+c=a(x−x
1
)(x−x
2
). Substitute
2
−3−
29
for x
1
and
2
−3+
29
for x
2
.
−x
2
−3x+5=−(x−
2
−
29
−3
)(x−
2
29
−3
)
EVALUATE
5−3x−x
2
Step-by-step explanation:
×
=
