Question

a. Factorise the expression f (x) = 2 x 3 – 7x2-3x + 18. Hence find all possible values of x for whichf(x)=0
​

Answer 1

Answer:

2×3-7×2-3x+18=0

6 - 14 - 3x + 18 = 0

-8 + 18 - 3x = 0

10 - 3x = 0

-3x = -10

x = 10/3.

Answer 2

Answer:

x=2,3 or −3/2

Step-by-step explanation:

Let f(x)=2x3–7x2–3x+18

Let f(x)=2x3–7x2–3x+18For x=2, the value of f(x) will be

Let f(x)=2x3–7x2–3x+18For x=2, the value of f(x) will bef(2)=2(2)3–7(2)2–3(2)+18

Let f(x)=2x3–7x2–3x+18For x=2, the value of f(x) will bef(2)=2(2)3–7(2)2–3(2)+18=16–28–6+18=0

Let f(x)=2x3–7x2–3x+18For x=2, the value of f(x) will bef(2)=2(2)3–7(2)2–3(2)+18=16–28–6+18=0As f(2)=0,(x–2) is a factor of f(x).

Let f(x)=2x3–7x2–3x+18For x=2, the value of f(x) will bef(2)=2(2)3–7(2)2–3(2)+18=16–28–6+18=0As f(2)=0,(x–2) is a factor of f(x).Now, performing long division we have

Let f(x)=2x3–7x2–3x+18For x=2, the value of f(x) will bef(2)=2(2)3–7(2)2–3(2)+18=16–28–6+18=0As f(2)=0,(x–2) is a factor of f(x).Now, performing long division we haveThus, f(x)=(x–2)(2x2–3x–9)

Let f(x)=2x3–7x2–3x+18For x=2, the value of f(x) will bef(2)=2(2)3–7(2)2–3(2)+18=16–28–6+18=0As f(2)=0,(x–2) is a factor of f(x).Now, performing long division we haveThus, f(x)=(x–2)(2x2–3x–9)=(x–2)(2x2–6x+3x–9)

Let f(x)=2x3–7x2–3x+18For x=2, the value of f(x) will bef(2)=2(2)3–7(2)2–3(2)+18=16–28–6+18=0As f(2)=0,(x–2) is a factor of f(x).Now, performing long division we haveThus, f(x)=(x–2)(2x2–3x–9)=(x–2)(2x2–6x+3x–9)=(x–2)[2x(x–3)+3(x–3)]

Let f(x)=2x3–7x2–3x+18For x=2, the value of f(x) will bef(2)=2(2)3–7(2)2–3(2)+18=16–28–6+18=0As f(2)=0,(x–2) is a factor of f(x).Now, performing long division we haveThus, f(x)=(x–2)(2x2–3x–9)=(x–2)(2x2–6x+3x–9)=(x–2)[2x(x–3)+3(x–3)]=(x–2)(x–3)(2x+3)

Let f(x)=2x3–7x2–3x+18For x=2, the value of f(x) will bef(2)=2(2)3–7(2)2–3(2)+18=16–28–6+18=0As f(2)=0,(x–2) is a factor of f(x).Now, performing long division we haveThus, f(x)=(x–2)(2x2–3x–9)=(x–2)(2x2–6x+3x–9)=(x–2)[2x(x–3)+3(x–3)]=(x–2)(x–3)(2x+3)Now, for f(x)=0

Let f(x)=2x3–7x2–3x+18For x=2, the value of f(x) will bef(2)=2(2)3–7(2)2–3(2)+18=16–28–6+18=0As f(2)=0,(x–2) is a factor of f(x).Now, performing long division we haveThus, f(x)=(x–2)(2x2–3x–9)=(x–2)(2x2–6x+3x–9)=(x–2)[2x(x–3)+3(x–3)]=(x–2)(x–3)(2x+3)Now, for f(x)=0(x–2)(x–3)(2x+3)=0

Let f(x)=2x3–7x2–3x+18For x=2, the value of f(x) will bef(2)=2(2)3–7(2)2–3(2)+18=16–28–6+18=0As f(2)=0,(x–2) is a factor of f(x).Now, performing long division we haveThus, f(x)=(x–2)(2x2–3x–9)=(x–2)(2x2–6x+3x–9)=(x–2)[2x(x–3)+3(x–3)]=(x–2)(x–3)(2x+3)Now, for f(x)=0(x–2)(x–3)(2x+3)=0Hence x=2,3 or −3/2