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Factorize:

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Concepts

Factorisation, Grouping terms, Factor theorem, Synthetic division

Explanation

We are given a cubic polynomial 3×3+5×2−16x−2. To factorize a cubic polynomial, we can try to find at least one root through trial, and then factor out the corresponding linear factor. What remains is a quadratic, which can be factorized further if possible.

Step-By-Step Solution

Step 1

Group terms to see if we can factor by grouping: 3×3+5×2−16x−2=(3×3+5×2)+(−16x−2)

Step 2

Factor out common terms from each group:

From 3×3+5×2, factor x2: 3×3+5×2=x2(3x+5)

From −16x−2, factor −2: −16x−2=−2(8x+1)

But 3x+5 and 8x+1 are not common, so grouping doesn’t immediately help. Let’s use the Rational Root Theorem and try likely rational roots (±1, ±2):

Step 3

Test x=−2: 3(−2)3+5(−2)2−16(−2)−2=3(−8)+5(4)+32−2=−24+20+32−2=26 Not zero.

Test x=2: 3(8)+5(4)−16(2)−2=24+20−32−2=44−34=10 Not zero.

Test x=1: 3(1)+5(1)−16(1)−2=3+5−16−2=8−18=−10 Not zero.

Test x=−1: 3(−1)+5(1)−16(−1)−2=−3+5+16−2=(2+16)−2=16 Not zero.

Test x=31​: 3(31​)3+5(31​)2−16(31​)−2=3(271​)+5(91​)−316​−2 =91​+95​−316​−2=96​−316​−2=32​−316​−2=−314​−2=−314​−36​=−320​ Not zero.

Test x=−31​: 3(−31​)3+5(−31​)2−16(−31​)−2=3(−271​)+5(91​)+316​−2 =−91​+95​+316​−2=94​+316​−2=94​+948​−918​=94+48−18​=934​=0

Test x=2: Already checked. Not zero.

Test x=−2: Already checked. Not zero.

Test x=11​=1, −1, already checked. Not zero.

Test x=32​: 3(32​)3+5(32​)2−16(32​)−2 (frac{2}{3})^3 = frac{8}{27}, (frac{2}{3})^2 = frac{4}{9}$ So 3×278​=2724​=98​ 5×94​=920​ 16×32​=332​

Sum: 98​+920​−332​−2 =928​−996​−918​ =928−96−18​=9−86​

None of the rational roots are zero. Let’s try factoring by splitting the middle term (grouping):

3×3+5×2−16x−2 See if we can write 5×2 as the sum of two terms so that grouping works. Break 5×2 as 6×2−x2: 3×3+6×2−x2−16x−2 Now, group: (3×3+6×2)+(−x2−16x−2) Take 3×2 from first, −(x2+16x+2) Not helping.

Alternatively, let’s pair differently: 3×3−16x+5×2−2=(3×3−16x)+(5×2−2) =x(3×2−16)+1(5×2−2) No common factor.

Alternatively, use the cubic factorisation formula: Cubic ax3+bx2+cx+d factors as (x+p)(ax2+qx+r) where p is a root.

Alternatively, use factor theorem and polynomial long division. Let’s try rational factor based on factorisation by splitting into (ax+b)(cx2+dx+e). Assume: 3×3+5×2−16x−2=(x+a)(3×2+bx+c)

Expand (x+a)(3×2+bx+c): =x(3×2+bx+c)+a(3×2+bx+c) =3×3+bx2+cx+3ax2+abx+ac

Collect like terms: 3×3+(b+3a)x2+(c+ab)x+ac

Set equal to 3×3+5×2−16x−2: So,

  • x3: 3
  • x2: b+3a=5
  • x1: c+ab=−16
  • x0: ac=−2

Now solve: From ac=−2, possible pairs are (a=1,c=−2) or (a=−1,c=2) or (a=2,c=−1) or (a=−2,c=1)

Try a=1,c=−2: Now, b+3a=5 ⇒ b=2 c+ab=−16 c=−2, a=1, b=2 −2+(1)(2)=−2+2=0=−16

Try a=−1,c=2: b +3(−1)=5, b−3=5⇒b=8 c + a b = 2 + (-1)(8) = 2 – 8 = -6 ne -16

Try a=2,c=−1: b +3(2)=5, b+6=5⇒b=−1 c + a b = -1 + 2(-1) = -1 -2 = -3 ne -16

Try a=−2,c=1: b +3(−2)=5, b−6=5⇒b=11 c + a b = 1 + (-2)(11) = 1 – 22 = -21 ne -16

Alternate factorization: a=2,c=−1, b=−1 So check if a=2, b=−1, c=−1 (But then ac=2×−1=−2, so possible.)

Test: c+ab=−1+2×(−1)=−1−2=−3=−16

Try other arrangements? Since nothing is working, perhaps this cubic is irreducible over rationals. Let’s use synthetic division with x+2:

Coefficients: 3 5 −16 −2 Test x=−2 (i.e., factor x+2):

Carry down 3. 3×−2=−6, 5−6=−1 −1×−2=2, −16+2=−14 −14×−2=28, −2+28=26 So remainder is 26=0.

Try x=2: 3 3×2=6, 5+6=11 11×2=22, −16+22=6 6×2=12, −2+12=10 Remainder 10=0

Try x=−1: 3 3×−1=−3, 5−3=2 2×−1=−2, −16−2=−18 −18×−1=18, −2+18=16 Remainder 16=0

Try x=1: 3 3×1=3, 5+3=8 8×1=8, −16+8=−8 −8×1=−8, −2−8=−10 Remainder −10=0

Now try x=2/3: 3 3×(2/3)=2, 5+2=7 7×(2/3)=14/3, −16+14/3=−48/3+14/3=−34/3 −34/3×(2/3)=−68/9, −2+(−68/9)=−18/9−68/9=−86/9 Not zero.

Try x=−2/3: 3 3×(−2/3)=−2, 5−2=3 3×(−2/3)=−2, −16−2=−18 −18×(−2/3)=12, −2+12=10 Remainder 10=0

At this point, let’s try using the quadratic formula to factor the cubic as a product of a linear and an irreducible quadratic (over reals):

Let us try to factor by pairs: 3×3+5×2−16x−2 Let’s factor it as (ax+b)(cx2+dx+e) Assume a=1 (standard), b=p

Let 3×3+5×2−16x−2=(x+p)(3×2+qx+r) Then, from above matching coefficients, as above:

From previous, a solution did not exist over rationals.

So, if we allow factorisation over real numbers (i.e., roots in surds):

Divide 3×3+5×2−16x−2 by x+p using synthetic division (or polynomial division), set p to be any root of f(x)=0.

Let’s use the cubic formula or numerically approximate one real root: According to Rational Root Theorem, there are no rational roots.

Use approximation; try x=−0.1: 3(−0.001)+5(0.01)−16(−0.1)−2=−0.003+0.05+1.6−2=(0.05+1.6)−2.003=1.65−2.003=−0.353

Try x=−0.2: 3(−0.008)+5(0.04)−16(−0.2)−2=−0.024+0.2+3.2−2=(0.2+3.2)−2.024=3.4−2.024=1.376

So root between x=−0.2 (1.376) and x=−0.1 (−0.353), so root between these two values.

By interpolation, let x≈−0.13.

But for Class 9 Assam Board, the expectation is to express as a product of a linear and an irreducible quadratic:

Therefore, 3×3+5×2−16x−2 cannot be factorized into rational polynomials.

But for academic needs, the cubic can be written as the product of a linear factor and a quadratic factor with irrational (real) coefficients: Let one real root be x=α (let’s estimate to 2 decimal places: x=−0.12 gives f(−0.12)=3(−0.001728)+5(0.0144)−16(−0.12)−2=−0.0051+0.072+1.92−2=1.987−2=−0.013

So at x=−0.12, value is −0.013; at x=−0.13, f(−0.13)=…

Alternatively, leave your answer as:

3×3+5×2−16x−2 is not factorisable over rational numbers.

Final Answer

3×3+5×2−16x−2​ cannot be factorized over rational numbers.

If you need to factorise over irrationals, you can express as 3×3+5×2−16x−2=(x+α)(ax2+bx+c), where α is the (approximate) real root x≈−0.12. But for Assam Board, leave as above.