# By the software Equation Grapher, we can find the smallest positive root of y=0 and then factorize the equation to find the other roots.

1.(a) By the software Equation Grapher, we can find the smallest positive root of y=0 and then factorize the equation to find the other roots.

(i)        From the graph, we can see that the smallest positive root is 1

f (1)        = 13 – 12 – 4(1) + 4

= 0

By the Remainder Theorem, 1 is a root and (x–1) is a factor of y = x3– x 2– 4x + 4

f (x)        = x3– x 2– 4x + 4

= (x–1) (x2–4)

= (x–1) (x–2) (x+2)

Therefore, by the Factor Theorem, the equation y = x3– x 2– 4x + 4 has roots at

x = -2, 1 and 2.

Hence, the smallest positive root is 1 or

(ii)        From the graph, it seems that the smallest positive root is

f () = 2() 3 – () 2 – 8() + 4

= 0

By the Remainder Theorem,  is root and (2x-1) is a factor of y = 2x3-x 2-8x+4

f(x)        = 2x3-x 2-8x+4

= (2x-1) (x2-4)

= (2x-1) (x-2) (x+2)

Therefore, by the Factor Theorem, the equation y = 2x3-x 2-8x+4 has roots at

x = -2,  and 2.

Hence, the smallest positive root is .

(iii)        From the graph, it seems that the smallest positive root is

f () = 3() 3 – () 2 – 12() + 4

= 0

By the Remainder Theorem,  is a root and (3x-1) is a factor of

y = 3x3-x 2-12x+4

f (x)        = 3x3-x 2-12x+4

= (3x-1) (x2-4)

= (3x-1) (x-2) (x+2)

Therefore, by the Factor Theorem, the equation y = 3x3-x 2-12x+4 has roots at

x = -2,  and 2.

Hence, the smallest positive root is .

(iv)        From the graph, it seems that the smallest positive root is

f () = 4() 3 – () 2 – 16() + 4

= 0

By the Remainder Theorem,  is a root and (4x-1) is a factor of

y = 4x3-x 2-16x+4

f (x)        = 4x3-x 2-16x+4

= (4x-1) (x2-4)

= (4x-1) (x-2) (x+2)

Therefore, by the Factor Theorem, the equation in (ii) has roots at

x = -2,  and 2.

Hence, the smallest positive root is .

1.(b)

From the graph, it seems that the smallest positive root is

f () =() 3 – () 2 () + 4

= 0

Hence, the smallest ...