To convert the standard form to factored form, we should know about about how to factorize the quadratic polynomial.
To factorize the quadratic polynomial which is in the form ax^{2} + bx + c, first we will check the coefficient of x^{2.}
Step 1 :
Step 2 :
Such that the product of those two factors is equal to c or ac and the when we combine it we should get the middle term (b).
Step 3 :
Write the middle term as sum of those two factors that we have received from step 2.
Step 4 :
Using grouping, write them as product of factors.
Problem 1 :
Convert f(x) = 3x^{2} + 18x - 48 to factored form.
Solution :
f(x) = 3x^{2} + 18x - 48
Here every term is a multiple of 3. Factoring 3 out.
f(x) = 3(x^{2} + 6x - 16)
-16 = -2⋅8
f(x) = 3(x^{2} -2x + 8x - 16)
f(x) = 3[x(x -2) + 8(x - 2)]
f(x) = 3(x - 2) (x + 8)
Problem 2 :
Factorize 2x^{2} +4x + 2
Solution :
Let f(x) = 2x^{2} +4x + 2
f(x) = 2(x^{2} + 2x + 1)
f(x) = 2(x^{2} + x + x + 1)
f(x) = 2[x(x + 1) + (x + 1)]
f(x) = 2 (x + 1) (x + 1)
Problem 3 :
The height of the foot ball kicked from the ground is given by the function h(t) = -5t^{2} + 20, where h(t) is height in meters and t is the time in seconds from its release.
(i) Write the function in factored form.
(ii) When will the foot ball hit the ground ?
Solution :
(i) h(t) = -5t^{2} + 20
Factoring -5, we get
h(t) = -5(t^{2} - 4)
h(t) = -5(t^{2} - 2^{2})
t^{2} - 2^{2 }looks like a^{2} - b^{2}
The expansion of a^{2} - b^{2 }= (a + b)(a - b)
h(t) = -5(t + 2) (t - 2)
(ii) When it hits the ground, the height will become zero.
0 = -5(t + 2) (t - 2)
Equating each factor to zero, we get
t + 2 = 0 and t - 2 = 0
t = -2 and t = 2
Problem 4 :
Solve x^{2} - 8x + 12 = -3
Move everything to one side and find zeros.
Solution :
x^{2} - 8x + 12 = -3
Add 3 on both sides, we get
x^{2} - 8x + 12 + 3 = 0
x^{2} - 8x + 15 = 0
x^{2} - 3x - 5x + 15 = 0
Using grouping, we get
x(x - 3) - 5(x - 3) = 0
(x - 5)(x - 3) = 0
Equating each factor to zero, we get
x - 5 = 0 and x - 3 = 0
x = 5 and x = 3
Problem 5 :
Find the points of intersections of the graphs
f(x) = x^{2} - 8x + 12 and g(x) = -3
Solution :
f(x) = x^{2} - 8x + 12 -----(1)
g(x) = -3 -----(2)
(1) = (2)
x^{2} - 8x + 12 = -3
Add 3 on both sides on both sides.
x^{2} - 8x + 12 + 3 = 0
x^{2} - 8x + 15 = 0
x^{2} - 3x - 5x + 15 = 0
Using grouping, we get
x(x - 3) - 5(x - 3) = 0
(x - 5)(x - 3) = 0
Equating each factor to zero, we get
x = 5 and x = 3
So, points of intersections are (5, -3) and (3, -3).
Problem 6 :
The path a dolphin travels when it rises above the ocean’s surface can be modelled by the function
h(d) = -0.2d^{2} + 2d
where h(d) is the height of the dolphin above the water’s surface and d is the horizontal distance from the point where the dolphin broke the water’s surface, both in feet. When will the dolphin reach a height of 1.8 feet?
Solution :
h(d) = -0.2d^{2} + 2d
When h(d) = 1.8
1.8 = -0.2d^{2} + 2d
-0.2d^{2} + 2d = 1.8
-0.2d^{2} + 2d - 1.8 = 0
Multiply the equation by -10, we get
2d^{2} + 20d + 18 = 0
Dividing by 2, we get
d^{2} - 10d + 9 = 0
(d - 1) (d - 9) = 0
Equating each factor to zero.
d - 1 = 0 and d - 9 = 0
d = 1 and d = 9
So, at d = 1 and d = 9 the dolphin will reach the height 200 m.
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